CFAI 2-Quantitative Methods Flashcards
Diferenças a calcular FV com annuity Due e com uma anuidade ordinária “ “
Quando resolvemos um problema com annuity due temos que ter em atenção se o cálculo de FV começasse em t-1, a fórmula de FV normal para n períodos considera que as anuidades são ordinárias! Portanto se quiseremos fazer 10 períodos em annuity due temos que fazer um FV normal que nos dá até ao 9º e posteriormente FV =PVt=9*(1+r)^1 e temos o 10º
FV=
PV(1+r)^n
PV=
FV/(1+r)^n
Growth rate or R formulae=
((FV/PV)^(1/n))-1
if the loan is needed for a $300,000 home and they tell you that the down payment is $50,000, make sure to…
reduce the amount borrowed, or PV, to $250,000! Plenty of folks will just grab the $300,000 number and plug it into the financial calculator.
when we have different ammounts of cash flows and we want to determine a FV or a PV the best way to do it is…
to separate de cash flows and resolve them separatly
NPV definição
Valor presente dos cash inflow menos o valor presente dos outflow. Serve para comparar projectos de investimento. Só se deve considerar investimento caso seja positivo.
WACC
Custo médio do capital para uma empresa. Ver fórmula no curso de Michigan
The Internal Rate of Return
is defined as the discount rate that makes NPV = 0.
Qual é o principal problema to IRR????
Favorece a pequena escala, projectos que podem criar mais valor em termos absolutos podem apresentar um IRR menor que projectos pequenos com pouca criação de valor. O IRR também favorece projetos que embora possam ter um NPV mais baixo pagam mais cedo. Logo o NPV reflete melhor o potencial do projecto.
Money-Weighted Rate of Return
A money-weighted rate of return is identical in concept to an internal rate of return: it is the discount rate on which the NPV = 0 or the present value of inflows = present value of outflows. Recall that for the IRR method, we start by identifying all cash inflows and outflows.
Example:
Each inflow or outflow must be discounted back to the present using a rate (r) that will make PV (inflows) = PV (outflows). For example, take a case where we buy one share of a stock for $50 that pays an annual $2 dividend, and sell it after two years for $65. Our money-weighted rate of return will be a rate that satisfies the following equation:
PV Outflows = PV Inflows = $2/(1 + r) + $2/(1 + r)2 + $65/(1 + r)2 = $50
Solving for r using a spreadsheet or financial calculator, we have a money-weighted rate of return = 17.78%.
LIMITATIONS
It’s important to understand the main limitation of the money-weighted return as a tool for evaluating managers. As defined earlier, the money-weighted rate of return factors all cash flows, including contributions and withdrawals. Assuming a money-weighted return is calculated over many periods, the formula will tend to place a greater weight on the performance in periods when the account size is highest (hence the label money-weighted).
Time-Weighted Rate of Return
The time-weighted rate of return is the preferred industry standard as it is not sensitive to contributions or withdrawals. It is defined as the compounded growth rate of $1 over the period being measured. The time-weighted formula is essentially a geometric mean of a number of holding-period returns that are linked together or compounded over time (thus, time-weighted).
The holding-period return, or HPR, (rate of return for one period) is computed using this formula:
HPR = ((MV1 - MV0 + D1 - CF1)/MV0)
Where: MV0 = beginning market value, MV1 = ending market value,
D1 = dividend/interest inflows, CF1 = cash flow received at period end (deposits subtracted, withdrawals added back)
compounded time-weighted rate of return, for N holding periods
= [(1 + HPR1)(1 + HPR2)(1 + HPR3) … *(1 + HPRN)] - 1.
Para cupões 0 a 1 ano para calcular a Discount Yield temos:
D/F*360/T=RBD
RBD - Bank Discount Yield
D = Diferença entre o face value e o preço atual
F = Face Value
T= nº de dias para a maturidade
360 nº de dias que se considera para o ano.
Neste tipo de exercicios pode pedir para resolver em ordem à yield ou dar a yield e pedir o preço p.e.
Effective Annual Yield, anualiza o Holding Period Yield (HPY ou HPReturn) de forma a permitir a comparação com outros investimentos é calculado por
EAY= (1+HPY)^365/t -1
Porque é que o Effective Annual Yield é maior que o Banks Discount yield
Remember that EAY > bank discount yield, for three reasons: (a) yield is based on purchase price, not face value, (b) it is annualized with compound interest (interest on interest), not simple interest, and (c) it is based on a 365-day year rather than 360 days. Be prepared to compare these two measures of yield and use these three reasons to explain why EAY is preferable.
Money Market Yield pode ser calculada em função da bank discount yield e da holding period Yield
Time wheigted yield
1:
HPYield anualizada de um periodo de 360 dias
(HPR)*(360/t)
2:
rMM = (360* rBD)/(360 - (t* rBD)
time weigthed é a média geométrica
o que é a Bond equivalent yield???
is simply the yield stated on a semiannual basis multiplied by 2. Thus, if you are given a semiannual yield of 3% and asked for the bond equivalent yield, the answer is 6%.
SKEW e Kurtosis o que são e como são interpretadas?
Skew, or skewness, can be mathematically defined as the averaged cubed deviation from the mean divided by the standard deviation cubed. If the result of the computation is greater than zero, the distribution is positively skewed. If it’s less than zero, it’s negatively skewed and equal to zero means it’s symmetric. For interpretation and analysis, focus on downside risk. Negatively skewed distributions have what statisticians call a long left tail (refer to graphs on previous page), which for investors can mean a greater chance of extremely negative outcomes. Positive skew would mean frequent small negative outcomes, and extremely bad scenarios are not as likely.
A nonsymmetrical or skewed distribution occurs when one side of the distribution does not mirror the other. Applied to investment returns, nonsymmetrical distributions are generally described as being either positively skewed (meaning frequent small losses and a few extreme gains) or negatively skewed (meaning frequent small gains and a few extreme losses).
or positively skewed distributions, the mode (point at the top of the curve) is less than the median (the point where 50% are above/50% below), which is less than the arithmetic mean (sum of observations/number of observations). The opposite rules apply to negatively skewed distribution: mode is greater than median, which is greater than arithmetic mean.
Positive: Mean > Median > Mode Negative: Mean
Diferentes tipos de Probabilidades (saber distinguir para o exame)
Empirical Probabilities
Empirical probabilities are objectively drawn from historical data. If we assembled a return distribution based on the past 20 years of data, and then used that same distribution to make forecasts, we have used an empirical approach. Of course, we know that past performance does not guarantee future results, so a purely empirical approach has its drawbacks.
Subjective Probabilities
Relationships must be stable for empirical probabilities to be accurate and for investments and the economy, relationships change. Thus, subjective probabilities are calculated; these draw upon experience and judgment to make forecasts or modify the probabilities indicated from a purely empirical approach. Of course, subjective probabilities are unique to the person making them and depend on his or her talents - the investment world is filled with people making incorrect subjective judgments.
A Priori Probabilities
A priori probabilities represent probabilities that are objective and based on deduction and reasoning about a particular case. For example, if we forecast that a company is 70% likely to win a bid on a contract (based on an either empirical or subjective approach), and we know this firm has just one business competitor, then we can also make an a priori forecast that there is a 30% probability that the bid will go to the competitor.
Joint probability o que é e como é calculada?
É a probabilidade de ocorrerem os dois eventos ao mesmo tempo.
Probability definitions can find their way into CFA exam questions. Naturally, there may also be questions that test the ability to calculate joint probabilities. Such computations require use of the multiplication rule, which states that the joint probability of A and B is the product of the conditional probability of A given B, times the probability of B. In probability notation:
Formula 2.20
Multiplication rule: P(AB) = P(A | B) * P(B)
Given a conditional probability P(A | B) = 40%, and a probability of B = 60%, the joint probability P(AB) = 0.6*0.4 or 24%, found by applying the multiplication rule.
Simplificando a probabilidade joint. Tendo a condicional P(A|B) se já temos a P(A) caso ocorra B, só temos que muultiplicar pela P(B)
Se forem independentes P(AB)= P(A) *P(B)
The Total Probability Rule
The total probability rule explains an unconditional probability of an event, in terms of that event’s conditional probabilities in a series of mutually exclusive, exhaustive scenarios. For the simplest example, there are two scenarios, S and the complement of S, or SC, and P(S) + P(SC) = 1, given the properties of being mutually exclusive and exhaustive. How do these two scenarios affect event A? P(A | S) and P(A | SC) are the conditional probabilities that event A will occur in scenario S and in scenario SC, respectively. If we know the conditional probabilities, and we know the probability of the two scenarios, we can use the total probability rule formula to find the probability of event A.
Formula 2.23
Total probability rule (two scenarios): P(A) = P(A | S)P(S) + P(A | SC)P(SC)
P(SC )- probabilidade “de não S”
correlação formula
covarAB/(desvpadA*desvpadB )
Bayes’ Formula Atualizar probabilidades caso ocorra um evento que a condicione
We all know intuitively of the principle that we learn from experience. For an analyst, learning from experience takes the form of adjusting expectations (and probability estimates) based on new information. Bayes’ formula essentially takes this principle and applies it to the probability concepts we have already learned, by showing how to calculate an updated probability, the new probability given this new information. Bayes’ formula is the updated probability, given new information:
Bayes’ Formula:
Conditional probability of new info. given the event * (Prior probability of the event)
Unconditional Probability of New Info
Formula 2.26
P(E | I) = P(I | E) / P(I) * P(E) Where: E = event, I = new info
Como tiramos as combinações possíveis para uma dada situação. P.e. quais as combinações possíveis de usar 5 trabalhadores para 5 diferentes postos de trabalho, etc?
Diferentes métodos
Method When appropriate?
Factorial Assigning a group of size n to n slots Combination Choosing r objects (in any order) from group of n Permutation Choosing r objects (in particular order) from group of n
The combination formula is used if the order of r does not matter. For choosing three objects from a total of five objects, we found 5!/(5 - 3)!*3!, or 10 ways.
The permutation formula is used if the order of r does matter. For choosing three objects from a total of five objects, we found 5!/(5 - 3)!, or 60 ways.
Factorial Notation
n! = n(n - 1)(n - 2) … 1. In other words, 5!, or 5 factorial is equal to (5)(4)(3)(2)*(1) = 120. In counting problems, it is used when there is a given group of size n, and the exercise is to assign the group to n slots; then the number of ways these assignments could be made is given by n!. If we were managing five employees and had five job functions, the number of possible combinations is 5! = 120.
Intervalos de confiança padrão, 1 desvio padrão, 2 desvios padrões e 3 desvios padrões
by assuming normal distribution, we are 68,3% confident that a variable will fall within one standard deviation. Within two standard deviation intervals, our confidence grows to 95,5%. Within three standard deviations, 99,7%.
Roy’s Safety-First Ratio
An optimal portfolio is one that minimizes the probability that the portfolio’s return will fall below a threshold level. In probability notation, if RP is the return on the portfolio, and RL is the threshold (the minimum acceptable return), then the portfolio for which P(RP
Q. A portfolio returned 5% over one year, if continuously compounded, this is equivalent to ____?
A. ln 5
B. ln 1.05
C. e5
D. e1.05
The answer would be B based on the definition of continuous compounding. A financial function calculator or spreadsheet could yield the actual percentage of 4.879%, but wouldn’t be necessary to answer the question correctly on the exam.
Exactly how large is large in terms of creating a large sample? De forma a termos uma distribuição normal…
Remember the number 30. According to the reference text, that’s the minimum number a sample must be before we can assume it is normally distributed. Don’t be surprised if a question asks how large a sample should be - should it be 20, 30, 40, or 50? It’s an easy way to test whether you’ve read the textbook, and if you remember 30, you score an easy correct answer.
Annuity:
FV e PV formulas
FV= A*([(1+r)^n-1]/r)
PV = A*((1-x)/r)
x= 1/(1+r)^n
Suppose your company’s defined contribution retirement plan allows you to invest up to €20,000 per year. You plan to invest €20,000 per year in a stock index fund for the next 30 years. Historically, this fund has earned 9 percent per year on average. Assuming that you actually earn 9 percent a year, how much money will you have available for retirement after making the last payment?
A = €20,000
r = 9% = 0.09
N = 30
FV annuity factor = (1+r)N−1r=(1.09)30−10.09=136.307539
FVN = €20,000(136.307539)
= €2,726,150.77
Assuming the fund continues to earn an average of 9 percent per year, you will have €2,726,150.77 available at retirement.
The manager of a Canadian pension fund knows that the fund must make a lump-sum payment of C$5 million 10 years from now. She wants to invest an amount today in a GIC so that it will grow to the required amount. The current interest rate on GICs is 6 percent a year, compounded monthly. How much should she invest today in the GIC?
Use Equation 9 to find the required present value:
FVN=C$5,000,000rs=6%=0.06m=12rs/m=0.06/12=0.005N=10mN=12(10)=120PV=FVN(1+rsm)−mN=C$5,000,000(1.005)−120=C$5,000,000(0.549633)=C$2,748,163.67
In applying Equation 9, we use the periodic rate (in this case, the monthly rate) and the appropriate number of periods with monthly compounding (in this case, 10 years of monthly compounding, or 120 periods).
Suppose you are considering purchasing a financial asset that promises to pay €1,000 per year for five years, with the first payment one year from now. The required rate of return is 12 percent per year. How much should you pay for this asset?
To find the value of the financial asset, use the formula for the present value of an ordinary annuity given in Equation 11 with the following data:
A = €1,000
r = 12% = 0.12
N = 5
PV annuity…=
= €1,000(3.604776)
= €3,604.78
The series of cash flows of €1,000 per year for five years is currently worth €3,604.78 when discounted at 12 percent.
You are interested in determining how long it will take an investment of €10,000,000 to double in value. The current interest rate is 7 percent compounded annually. How many years will it take €10,000,000 to double to €20,000,000?
Use Equation 2, FVN = PV(1 + r)N, to solve for the number of periods, N, as follows:
(1+r)N=FVN/PV=2Nln(1+r)=ln(2)N=ln(2)/ln(1+r)=ln(2)/ln(1.07)=10.24
With an interest rate of 7 percent, it will take approximately 10 years for the initial €10,000,000 investment to grow to €20,000,000. Solving for N in the expression (1.07)N = 2.0 requires taking the natural logarithm of both sides and using the rule that ln(xN) = N ln(x). Generally, we find that N = [ln(FV/PV)]/ln(1 + r). Here, N = ln(€20,000,000/ €10,000,000)/ln(1.07) = ln(2)/ln(1.07) = 10.24
A bank quotes a rate of 5.89 percent with an effective annual rate of 6.05 percent. Does the bank use annual, quarterly, or monthly compounding?
For annual compounding, with m = 1, 1.0605 ≠ 1.0589.
For quarterly compounding, with m = 4, 1.0605 ≠ 1.060214.
For monthly compounding, with m = 12, 1.0605 ≈ 1.060516.
Two years from now, a client will receive the first of three annual payments of $20,000 from a small business project. If she can earn 9 percent annually on her investments and plans to retire in six years, how much will the three business project payments be worth at the time of her retirement?
In summary, your client will have $77,894.21 in six years if she receives three yearly payments of $20,000 starting in Year 2 and can earn 9 percent annually on her investments.
•A client has agreed to invest €100,000 one year from now in a business planning to expand, and she has decided to set aside the funds today in a bank account that pays 7 percent compounded quarterly. How much does she need to set aside?
Use your calculator’s financial functions to verify that the present value, X, equals €93,295.85.
In summary, your client will have to deposit €93,295.85 today to have €100,000 in one year if her bank account pays 7 percent compounded quarterly.
•A client can choose between receiving 10 annual $100,000 retirement payments, starting one year from today, or receiving a lump sum today. Knowing that he can invest at a rate of 5 percent annually, he has decided to take the lump sum. What lump sum today will be equivalent to the future annual payments?
In summary, the present value of 10 payments of $100,000 is $772,173.49 if the first payment is received in one year and the rate is 5 percent compounded annually. Your client should accept no less than this amount for his lump sum payment.
A perpetual preferred stock position pays quarterly dividends of $1,000 indefinitely (forever). If an investor has a required rate of return of 12 percent per year compounded quarterly on this type of investment, how much should he be willing to pay for this dividend stream?
The investor will have to pay $33,333.33 today to receive $1,000 per quarter forever if his required rate of return is 3 percent per quarter (12 percent per year).
•Suppose you plan to send your daughter to college in three years. You expect her to earn two-thirds of her tuition payment in scholarship money, so you estimate that your payments will be $10,000 a year for four years. To estimate whether you have set aside enough money, you ignore possible inflation in tuition payments and assume that you can earn 8 percent annually on your investments. How much should you set aside now to cover these payments?
In summary, you should set aside $28,396.15 today to cover four payments of $10,000 starting in three years if your investments earn a rate of 8 percent annually.
•A client is confused about two terms on some certificate-of-deposit rates quoted at his bank in the United States. You explain that the stated annual interest rate is an annual rate that does not take into account compounding within a year. The rate his bank calls APY (annual percentage yield) is the effective annual rate taking into account compounding. The bank’s customer service representative mentioned monthly compounding, with $1,000 becoming $1,061.68 at the end of a year. To prepare to explain the terms to your client, calculate the stated annual interest rate that the bank must be quoting.
Use your calculator’s financial functions to verify that the stated interest rate of the savings account is 6 percent with monthly compounding.
A client seeking liquidity sets aside €35,000 in a bank account today. The account pays 5 percent compounded monthly. Because the client is concerned about the fact that deposit insurance covers the account for only up to €100,000, calculate how many months it will take to reach that amount.
Use your calculator’s financial functions to verify that your client will have to wait 252.48 months to have €100,000 if he deposits €35,000 today in a bank account paying 5 percent compounded monthly. (Some calculators will give 253 months.)
A client plans to send a child to college for four years starting 18 years from now. Having set aside money for tuition, she decides to plan for room and board also. She estimates these costs at $20,000 per year, payable at the beginning of each year, by the time her child goes to college. If she starts next year and makes 17 payments into a savings account paying 5 percent annually, what annual payments must she make?
In summary, your client will have to save $2,744.50 each year if she starts next year and makes 17 payments into a savings account paying 5 percent annually.
•You are analyzing the last five years of earnings per share data for a company. The figures are $4.00, $4.50, $5.00, $6.00, and $7.00. At what compound annual rate did EPS grow during these years?
EPS grew at an annual rate of 15.02 percent during the four years.
•An analyst expects that a company’s net sales will double and the company’s net income will triple over the next five-year period starting now. Based on the analyst’s expectations, which of the following best describes the expected compound annual growth?
A.Net sales will grow 15% annually and net income will grow 25% annually.
B.Net sales will grow 20% annually and net income will grow 40% annually.
C.Net sales will grow 25% annually and net income will grow 50% annually.
A is correct. Using the general time value of money formula, for sales, solve for r in the equation 2 = 1 × (1 + r)5. For income, solve 3 = 1 × (1 + r)5. Alternatively, using a financial calculator, for sales, enter N = 5, PV = 1, PMT = 0, FV = −2 and compute I/Y. For income, change the FV to −3 and again solve for I/Y. The solution for sales is 14.87%; and for income is 24.57%.
John Wilson buys 150 shares of ABM on 1 January 2012 at a price of $156.30 per share. A dividend of $10 per share is paid on 1 January 2013. Assume that this dividend is not reinvested. Also on 1 January 2013, Wilson sells 100 shares at a price of $165 per share. On 1 January 2014, he collects a dividend of $15 per share (on 50 shares) and sells his remaining 50 shares at $170 per share.
A.Write the formula to calculate the money-weighted rate of return on Wilson’s portfolio.
B.Using any method, compute the money-weighted rate of return.
C.Calculate the time-weighted rate of return on Wilson’s portfolio.
D.Describe a set of circumstances for which the money-weighted rate of return is an appropriate return measure for Wilson’s portfolio.
E.Describe a set of circumstances for which the time-weighted rate of return is an appropriate return measure for Wilson’s portfolio.
A•The money-weighted rate of return is the discount rate that equates the present value of inflows to the present value of outflows.
Outflows:
Att=0(1January2012): 150sharespurchased×$156.30pershare=$23,445
Inflows:
Att=1(1January2013): 150shares×$10dividendpershare=$1,500 100sharessold×$165pershare=$16,500Att=2(1January2014): 50sharespurchased×$15dividendpershare=$750 50sharessold×$170pershare=$8,500
PV(Outflows)=PV(Inflows) 23,445=1,500+16,5001+r+750+8,500(1+r)2 =18,0001+r+9,250(1+r)2
The last line is the equation for calculating the money-weighted rate of return on Wilson’s portfolio.
B•We can solve for the money-weighted return by entering −23,445, 18,000, and 9,250 in a spreadsheet or calculator with an IRR function. In this case, we can also solve for money-weighted rate of return as the real root of the quadratic equation 18,000x + 9,250x2 − 23,445 = 0, where x = 1/(1 + r). By any method, the solution is r = 0.120017 or approximately 12 percent.
C•The time-weighted rate of return is the solution to (1 + Time-weighted rate of return)2 = (1 + r1)(1 + r2), where r1 and r2 are the holding period returns in the first and second years, respectively. The value of the portfolio at t = 0 is $23,445. At t = 1, there are inflows of sale proceeds of $16,500 and $1,500 in dividends, or $18,000 in total. The balance of 50 shares is worth $8,250 = 50 shares × $165 per share. So at t = 1 the valuation is $26,250 = $18,000 + $8,250. Thus
r1 = ($26,250 – $23,445)/$23,445 = 0.119642 for the first year
The amount invested at t = 1 is $8,250 = (50 shares)($165 per share). At t = 2, $750 in dividends are received, as well as sale proceeds of $8,500 (50 shares sold × $170 per share). So at t = 2, the valuation is $9,250 = $750 + $8,500. Thus
r2 = ($9,250 – $8,250)/$8,250 = 0.121212 for the second year
Time-weighted rate of return = (1.119642)(1.121212)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√−1=0.1204
or approximately 12 percent.
D•If Wilson is a private investor with full discretionary control over the timing and amount of withdrawals and additions to his portfolios, then the money-weighted rate of return is an appropriate measure of portfolio returns.
E•If Wilson is an investment manager whose clients exercise discretionary control over the timing and amount of withdrawals and additions to the portfolio, then the time-weighted rate of return is the appropriate measure of portfolio returns. Time-weighted rate of return is standard in the investment management industry.
•A Treasury bill with a face value of $100,000 and 120 days until maturity is selling for $98,500.
A.What is the T-bill’s bank discount yield?
B.What is the T-bill’s money market yield?
C.What is the T-bill’s effective annual yield?
In this solution, F stands for face value, P stands for price, and D stands for the discount from face value (D = F − P).
(A) .Use the discount yield formula (Equation 3), rBD = D/F × 360/t:
rBD=($1,500/$100,000)×(360/120)=0.0150×3=0.045
The T-bill’s bank discount yield is 4.5 percent a year.
(B). Use your answer from Part A and the money market yield formula (Equation 6), rMM = (360 × rBD)/(360 − t × rBD):
rMM=(360×0.045)/(360−120×0.045)=0.04568
The T-bill’s money market yield is 4.57 percent a year.
(C). Calculate the holding period yield (using Equation 4), then compound it forward to one year. First, the holding period yield (HPY) is
HPY=P1−P+D1P0=(100,000−98,500)/98,500=0.015228
Next, compound the 120-day holding period yield, a periodic rate, forward to one year using Equation 5:
Effectiveannualyield=(1+HPY)365/ t− 1Effectiveannualyield=(1.015228)365/ 120− 1=0.047044
The T-bill’s effective annual yield is 4.7 percent a year.
•Jane Cavell has just purchased a 90-day US Treasury bill. She is familiar with yield quotes on German Treasury discount paper but confused about the bank discount quoting convention for the US T-bill she just purchased.
A.Discuss three reasons why bank discount yield is not a meaningful measure of return.
B.Discuss the advantage of money market yield compared with bank discount yield as a measure of return.
C.Explain how the bank discount yield can be converted to an estimate of the holding period return Cavell can expect if she holds the T-bill to maturity.
- In the United States, T-bill yields are quoted on a bank discount basis. The bank discount yield is not a meaningful measure of the return for three reasons: First, the yield is based on the face value of the bond, not on its purchase price. Returns from investments should be evaluated relative to the amount that is invested. Second, the yield is annualized based on a 360-day year rather than a 365-day year. Third, the bank discount yield annualizes with simple interest, which ignores the opportunity to earn interest on interest (compound interest).
- The money market yield is superior to the bank discount yield because the money market yield is computed relative to the purchase price (not the face value).
- The T-bill yield can be restated on a money market basis by multiplying the bank discount yield by the ratio of the face value to the purchase price. Cavell could divide the annualized yield by 4 to compute the 90-day holding period yield. This is a more meaningful measure of the return that she will actually earn over 90 days (assuming that she holds the T-bill until it matures).
Medidores de dispersão
Range - valor máximo no intervalo menos mínimo
MAD = Mean Absolute Deviation Sum(Xi-Média)/N
Variancia = Sum (Xi-Média)^2/N
Semivariance= Sum (Yi-Média)^2/(n-1) Yi = valores que são inferiores à média p.e. para considerar apenas a variância negativa
Chebyshev’s Inequality
Definition of Chebyshev’s Inequality. According to Chebyshev’s inequality, for any distribution with finite variance, the proportion of the observations within k standard deviations of the arithmetic mean is at least
1 − 1/k^2 dá-nos a % mínima de observações dentro dos desvios padrões k
State the type of scale used to measure the following sets of data.
A.Sales in euros.
B.The investment style of mutual funds.
C.An analyst’s rating of a stock as underweight, market weight, or overweight, referring to the analyst’s suggested weighting of the stock in a portfolio.
D.A measure of the risk of portfolios on a scale of whole numbers from 1 (very conservative) to 5 (very risky) where the difference between 1 and 2 represents the same increment in risk as the difference between 4 and 5.
- Sales in euros are measured on a ratio scale.
- Mutual fund investment styles are measured on a nominal scale. We can count the number of funds following a particular style, but whatever classification scheme we use, we cannot order styles into “greater than” or “less than” relationships.
- The ratings are measured on an ordinal scale. An analyst’s rating of a stock as underweight, market weight, or overweight orders the rated securities in terms of levels of expected investment performance.
- The risk measurements are measured on an interval scale because not only do the measurements involve a ranking, but differences between adjacent values represent equal differences in risk. Because the measurement scale does not have a true zero, they are not measured on a ratio scale.
•If the observations in a data set have different values, is the geometric mean for that data set less than that data set’s:
harmonic mean? arithmetic mean?
A No No
B No Yes
C Yes No
•B is correct. Unless all the values of the observations in a data set have the same value, the harmonic mean is less than the corresponding geometric mean, which in turn is less than the corresponding arithmetic mean. In other words, regarding means, typically harmonic mean
•Is a return distribution characterized by frequent small losses and a few large gains best described as having:
negative skew? a mean that is greater than the median?A No No
B No Yes
C Yes No
•B is correct. A distribution with frequent small losses and a few large gains has positive skew (long tail on the right side) and the mean is greater than the median.
•An analyst gathered the following information about the return distributions for two portfolios during the same time period:
Skewness Kurtosis
Portfolio A −1.3 2.2
Portfolio B 0.5 3.5
The analyst stated that the distribution for Portfolio A is more peaked than a normal distribution and that the distribution for Portfolio B has a long tail on the left side of the distribution. Which of the following is most true?
A.The statement is not correct in reference to either portfolio.
B.The statement is correct in reference to Portfolio A, but the statement is not correct in reference to Portfolio B.
C.The statement is not correct in reference to Portfolio A, but the statement is correct in reference to Portfolio B.
•A is correct. The analyst’s statement is not correct in reference to either portfolio. Portfolio A has a kurtosis of less than 3 meaning that it is less peaked than a normal distribution (platykurtic). Portfolio B is positively skewed (long tail on the right side of the distribution).
•The coefficient of variation is useful in determining the relative degree of variability of different data sets if those data sets have different:
A.means or different units of measurement.
B.means, but not different units of measurement.
C.units of measurement, but not different means.
•A is correct. The coefficient of variation is a relative measure of risk (dispersion) and is useful for both data sets that have different means and for data sets that do not have the same unit of measurement.
•An analyst gathered the following information about a portfolio’s performance over the past ten years:
Mean annual return 11.8%
Standard deviation of annual returns 15.7%
Portfolio beta 1.2
If the mean return on the risk-free asset over the same period was 5.0%, the coefficient of variation and Sharpe ratio, respectively, for the portfolio are closest to:
Coefficient of variation Sharperatio A 0.75 0.43 B 1.33 0.36 C 1.33 0.43
C is correct. The coefficient of variation measures total risk per unit of return or standard deviation/mean return, or 15.7/11.8 = 1.33. The Sharpe ratio is excess return per unit of risk or excess return/standard deviation. The mean excess return is 11.8% − 5.0% = 6.8%, so the Sharpe ratio is 6.8/15.7 = 0.43.
Retorno esperado de uma obrigação com 0% recovery rate e sabendo as probabilidades de default, como se calcula?
Retorno esperado= Pdefault0+(1+r)(1-Pdefault)
Sendo que r é a taxa que a obrigação oferece.
para N ativos quantas covariancias podemos tirar?
[N*(N-1)]/2
Bayes’ formula makes use of Equation 6, the total probability rule. To review, that rule expressed the probability of an event as a weighted average of the probabilities of the event, given a set of scenarios. Bayes’ formula works in reverse; more precisely, it reverses the “given that” information. Bayes’ formula uses the occurrence of the event to infer the probability of the scenario generating it. For that reason, Bayes’ formula is sometimes called an inverse probability. In many applications, including the one illustrating its use in this section, an individual is updating his beliefs concerning the causes that may have produced a new observation.
You are still an investor in DriveMed stock. To review the givens, your prior probabilities are P(EPS exceeded consensus) = 0.45, P(EPS met consensus) = 0.30, and P(EPS fell short of consensus) = 0.25. You also have the following conditional probabilities:
P(DriveMed expands | EPS exceeded consensus) = 0.75
P(DriveMed expands | EPS met consensus) = 0.20
P(DriveMed expands | EPS fell short of consensus) = 0.05
Recall that you updated your probability that last quarter’s EPS exceeded the consensus estimate from 45 percent to 82.3 percent after DriveMed announced it would expand. Now you want to update your other priors.
- Update your prior probability that DriveMed’s EPS met consensus.
- Update your prior probability that DriveMed’s EPS fell short of consensus.
- Show that the three updated probabilities sum to 1. (Carry each probability to four decimal places.)
- Suppose, because of lack of prior beliefs about whether DriveMed would meet consensus, you updated on the basis of prior probabilities that all three possibilities were equally likely: P(EPS exceeded consensus) = P(EPS met consensus) = P(EPS fell short of consensus) = 1/3. What is your estimate of the probability P(EPS exceeded consensus | DriveMed expands)?
P(Event∣∣Information)=[P(Information∣∣Event)/P(Information)]*P(Event)
1
After taking account of the announcement on expansion, your updated probability that last quarter’s EPS for DriveMed just met consensus is 14.6 percent compared with your prior probability of 30 percent.
2
=(0.05/0.41)(0.25)=0.121951(0.25)=0.030488
As a result of the announcement, you have revised your probability that DriveMed’s EPS fell short of consensus from 25 percent (your prior probability) to 3 percent.
4
Not surprisingly, the probability of DriveMed expanding is 1/3 because the decision maker has no prior beliefs or views regarding how well EPS performed relative to the consensus estimate. Now we can use Bayes’ formula to find P(EPS exceeded consensus | DriveMed expands) = [P(DriveMed expands | EPS exceeded consensus)/P(DriveMed expands)] P(EPS exceeded consensus) = (0.75/(1/3) = 0.75 or 75 percent. This probability is identical to your estimate of P(DriveMed expands | EPS exceeded consensus).
We can illustrate this formula with the binomial option pricing model. This model describes the movement of the underlying asset as a series of moves, price up (U) or price down (D). For example, two sequences of five moves containing three up moves, such as UUUDD and UDUUD, result in the same final stock price. At least for an option with a payoff dependent on final stock price, the number but not the order of up moves in a sequence matters. How many sequences of five moves belong to the group with three up moves?
We can illustrate this formula with the binomial option pricing model. This model describes the movement of the underlying asset as a series of moves, price up (U) or price down (D). For example, two sequences of five moves containing three up moves, such as UUUDD and UDUUD, result in the same final stock price. At least for an option with a payoff dependent on final stock price, the number but not the order of up moves in a sequence matters. How many sequences of five moves belong to the group with three up moves? The answer is 10, calculated using the combination formula (“5 choose 3”):
5C3=5!/(5−3)!3!
A useful fact can be illustrated as follows: 5C3 = 5!/2!3! equals 5C2 = 5!/3!2!, as 3 + 2 = 5; 5C4 = 5!/1!4! equals 5C1 = 5!/4!1!, as 4 + 1 = 5. This symmetrical relationship can save work when we need to calculate many possible combinations.
Suppose jurors want to select three companies out of a group of five to receive the first-, second-, and third-place awards for the best annual report. In how many ways can the jurors make the three awards? Order does matter if we want to distinguish among the three awards (the rank within the group of three); clearly the question makes order important. On the other hand, if the question were “In how many ways can the jurors choose three winners, without regard to place of finish?” we would use the combination formula.
Suppose jurors want to select three companies out of a group of five to receive the first-, second-, and third-place awards for the best annual report. In how many ways can the jurors make the three awards? Order does matter if we want to distinguish among the three awards (the rank within the group of three); clearly the question makes order important. On the other hand, if the question were “In how many ways can the jurors choose three winners, without regard to place of finish?” we would use the combination formula.
To address the first question above, we need to count ordered listings such as first place, New Company; second place, Fir Company; third place, Well Company. An ordered listing is known as a permutation, and the formula that counts the number of permutations is known as the permutation formula.22
◾Permutation Formula. The number of ways that we can choose r objects from a total of n objects, when the order in which the r objects are listed does matter, is
nPr=n!(n−r)!
So the jurors have 5P3 = 5!/(5 − 3)! = (5)(4)(3)(2)(1)/(2)(1) = 120/2 = 60 ways in which they can make their awards.
Multinomial Formula (General Formula for Labeling Problems). The number of ways that n objects can be labeled with k different labels, with n1 of the first type, n2 of the second type, and so on, with n1 + n2 + … + nk = n, is given by
n!/(n1! n2!…nk!)
factorial formula
n!/(n1! n2!…nk!) É parecida à nPr contudo dá label a todas.
n! factorial dá-nos todas as opções.
As 5 diferentes maneiras de assignar amostras de um grupo.
factorial... ... ... .. ..
- The number of ways to assign every member of a group of size n to n slots is n! = n (n − 1) (n − 2)(n − 3) … 1. (By convention, 0! = 1.)
- The number of ways that n objects can be labeled with k different labels, with n1 of the first type, n2 of the second type, and so on, with n1 + n2 + … + nk = n, is given by n!/(n1!n2! … nk!). This expression is the multinomial formula.
- A special case of the multinomial formula is the combination formula. The number of ways to choose r objects from a total of n objects, when the order in which the r objects are listed does not matter, is
nCr=(nr)=n!(n−r)!r!
•The number of ways to choose r objects from a total of n objects, when the order in which the r objects are listed does matter, is
nPr=n!(n−r)!
This expression is the permutation formula.
Label each of the following as an empirical, a priori, or subjective probability.
A.The probability that US stock returns exceed long-term corporate bond returns over a 10-year period, based on Ibbotson Associates data.
B.An updated (posterior) probability of an event arrived at using Bayes’ formula and the perceived prior probability of the event.
C.The probability of a particular outcome when exactly 12 equally likely possible outcomes exist.
D.A historical probability of default for double-B rated bonds, adjusted to reflect your perceptions of changes in the quality of double-B rated issuance.
- The probability is an empirical probability.
- The probability is a subjective probability.
- The probability is an a priori probability.
- The probability is a subjective probability.
Suppose you have two limit orders outstanding on two different stocks. The probability that the first limit order executes before the close of trading is 0.45. The probability that the second limit order executes before the close of trading is 0.20. The probability that the two orders both execute before the close of trading is 0.10. What is the probability that at least one of the two limit orders executes before the close of trading?
The probability that at least one of the two orders executes is given by the addition rule for probabilities. Let A stand for the event that the first limit order executes before the close of trading [P(A) = 0.45] and let B stand for the event that the second limit order executes before the close of trading [P(B) = 0.20]. P(AB) is given as 0.10. Therefore, P(A or B) = P(A) + P(B) − P(AB) = 0.45 + 0.20 − 0.10 = 0.55. The probability that at least one of the two orders executes before the close of trading is 0.55.
You apply both valuation criteria and financial strength criteria in choosing stocks. The probability that a randomly selected stock (from your investment universe) meets your valuation criteria is 0.25. Given that a stock meets your valuation criteria, the probability that the stock meets your financial strength criteria is 0.40. What is the probability that a stock meets both your valuation and financial strength criteria?
Use Equation 2, the multiplication rule for probabilities P(AB) = P(A | B)P(B), defining A as the event that a stock meets the financial strength criteria and defining B as the event that a stock meets the valuation criteria. Then P(AB) = P(A | B)P(B) = 0.40 × 0.25 = 0.10. The probability that a stock meets both the financial and valuation criteria is 0.10.
A report from Fitch data service states the following two facts:1
◾In 2002, the volume of defaulted US high-yield debt was $109.8 billion. The average market size of the high-yield bond market during 2002 was $669.5 billion.
◾The average recovery rate for defaulted US high-yield bonds in 2002 (defined as average price one month after default) was $0.22 on the dollar.
Address the following three tasks:
A.On the basis of the first fact given above, calculate the default rate on US high-yield debt in 2002. Interpret this default rate as a probability.
B.State the probability computed in Part A as an odds against default.
C.The quantity 1 minus the recovery rate given in the second fact above is the expected loss per $1 of principal value, given that default has occurred. Suppose you are told that an institution held a diversified high-yield bond portfolio in 2002. Using the information in both facts, what was the institution’s expected loss in 2002, per $1 of principal value of the bond portfolio?
A.The default rate was ($109.8 billion)/($669.5 billion) = 0.164 or 16.4 percent. This result can be interpreted as the probability that $1 invested in a market-value-weighted portfolio of US high-yield bonds was subject to default in 2002.
B.The odds against an event are denoted E = [1 − P(E)]/P(E). In this case, the odds against default are (1 − 0.164)/0.164 = 5.098, or “5.1 to 1.”
C.First, note that E(loss | bond defaults) = 1 − $0.22 = $0.78. According to the total probability rule for expected value, E(loss) = E(loss | bond defaults)P(bond defaults) + E(loss | bond does not default)P(bond does not default) = ($0.78)(0.164) + ($0.0)(0.836) = 0.128, or $0.128. Thus, the institution’s expected loss was approximately 13 cents per dollar of principal value invested.
The variance of a stock portfolio depends on the variances of each individual stock in the portfolio and also the covariances among the stocks in the portfolio. If you have five stocks, how many unique covariances (excluding variances) must you use in order to compute the variance of return on your portfolio? (Recall that the covariance of a stock with itself is the stock’s variance.)
n*(n-1)/2
= 10
On one day in March, 3,292 issues traded on the NYSE: 1,303 advanced, 1,764 declined, and 225 were unchanged. In how many ways could this set of outcomes have happened? (Set up the problem but do not solve it.)
•This is a labeling problem in which we assign each NYSE issue a label: advanced, declined, or unchanged. The expression to count the number of ways 3,292 issues can be assigned to these three categories such that 1,303 advanced, 1,764 declined, and 225 remained unchanged is 3,292!/(1,303!)(1,764!)(225!).
Your firm intends to select 4 of 10 vice presidents for the investment committee. How many different groups of four are possible?
nCr
.
Here,n=10 and r=4,so the answer is 10!/[(10−4)!4!]=3,628,800/(720)(24)= 210.
You have developed a set of criteria for evaluating distressed credits. Companies that do not receive a passing score are classed as likely to go bankrupt within 12 months. You gathered the following information when validating the criteria:
◾Forty percent of the companies to which the test is administered will go bankrupt within 12 months: P(nonsurvivor) = 0.40.
◾Fifty-five percent of the companies to which the test is administered pass it: P(pass test) = 0.55.
◾The probability that a company will pass the test given that it will subsequently survive 12 months, is 0.85: P(pass test | survivor) = 0.85.
A.What is P(pass test | nonsurvivor)?
B.Using Bayes’ formula, calculate the probability that a company is a survivor, given that it passes the test; that is, calculate P(survivor | pass test).
C.What is the probability that a company is a nonsurvivor, given that it fails the test?
D.Is the test effective?
A •We can set up the equation using the total probability rule:
P(passtest)=P(passtest∣∣survivor)P(survivor)+P(passtest∣∣nonsurvivor)P(nonsurvivor)
WeknowthatP(survivor)=1−P(nonsurvivor)=1−0.40=0.60.Therefore,P(passtest)=0.55=0.85(0.60)+P(passtest∣∣nonsurvivor)(0.40).ThusP(passtest∣∣nonsurvivor)=[0.55−0.85(0.60)]/0.40=0.10.
B •P(survivor∣∣passtest)=[P(passtest∣∣survivor)/P(passtest)]P(survivor)=(0.85/0.55)0.60=0.927273
The information that a company passes the test causes you to update your probability that it is a survivor from 0.60 to approximately 0.927.
C •According to Bayes’ formula, P(nonsurvivor | fail test) = [P(fail test | nonsurvivor)/ P(fail test)]P(nonsurvivor) = [P(fail test | nonsurvivor)/0.45]0.40.
We can set up the following equation to obtain P(fail test | nonsurvivor):
P(failtest)=P(failtest∣∣nonsurvivor)P(nonsurvivor)+P(failtest∣∣survivor)P(survivor)0.45=P(failtest∣∣nonsurvivor)0.40+0.15(0.60)
where P(fail test | survivor) = 1 − P(pass test | survivor) = 1 − 0.85 = 0.15. So P(fail test | nonsurvivor) = [0.45 − 0.15(0.60)]/0.40 = 0.90. Using this result with the formula above, we find P(nonsurvivor | fail test) = (0.90/0.45)0.40 = 0.80. Seeing that a company fails the test causes us to update the probability that it is a nonsurvivor from 0.40 to 0.80. tilizar.
Central Limit Theorem
Qual é o nº mínimo da amostra?
Como se calcula a média e o desvio padrão da amostra?
•The distribution of the sample mean X
will be approximately normal.
•The mean of the distribution of X
will be equal to the mean of the population from which the samples are drawn.
•The variance of the distribution of X
will be equal to the variance of the population divided by the sample size.
Intervalos de confiaça com 90% de probabilidade 95% e 99%
Reliability Factors for Confidence Intervals Based on the Standard Normal Distribution. We use the following reliability factors when we construct confidence intervals based on the standard normal distribution:14
◾90 percent confidence intervals: Use z0.05 = 1.65
◾95 percent confidence intervals: Use z0.025 = 1.96
◾99 percent confidence intervals: Use z0.005 = 2.58
Média ±z(α/2)*σ/sqrt(n)
α/2 Alpha é o intervalo de confiaça os valores são retirados da tabela da normal.
Parecido com o cálculo do VAR para um dado intervalo, contudo os valores são diferentes.
90% equivale a 95% do Var porque o VAR apenas considera as quedas de preço.
Suppose an investment analyst takes a random sample of US equity mutual funds and calculates the average Sharpe ratio. The sample size is 100, and the average Sharpe ratio is 0.45. The sample has a standard deviation of 0.30. Calculate and interpret the 90 percent confidence interval for the population mean of all US equity mutual funds by using a reliability factor based on the standard normal distribution.
The reliability factor for a 90 percent confidence interval, as given earlier, is z0.05 = 1.65. The confidence interval will be
The confidence interval spans 0.4005 to 0.4995, or 0.40 to 0.50, carrying two decimal places. The analyst can say with 90 percent confidence that the interval includes the population mean.
In this example, the analyst makes no specific assumption about the probability distribution describing the population. Rather, the analyst relies on the central limit theorem to produce an approximate normal distribution for the sample mean.
t distribution
Of the three distributions shown in Figure 1, the standard normal distribution has tails that approach zero faster than the tails of the two t-distributions. The t-distribution is also symmetrically distributed around its mean value of zero, just like the normal distribution. As the degrees of freedom increase, the t-distribution approaches the standard normal. The t-distribution with df = 8 is closer to the standard normal than the t-distribution with df = 2.
Beyond plus and minus four standard deviations from the mean, the area under the standard normal distribution appears to approach 0; both t-distributions continue to show some area under each curve beyond four standard deviations, however. The t-distributions have fatter tails, but the tails of the t-distribution with df = 8 more closely resemble the normal distribution’s tails. As the degrees of freedom increase, the tails of the t-distribution become less fat.
Quando a amostra é pequena e não dá para usar uma distribuição normal, usa-se uma distribuiução t.student
- The degrees of freedom number for use with the t-distribution is also n − 1.
- The t-distribution has fatter tails than the standard normal distribution but converges to the standard normal distribution as degrees of freedom go to infinity.
Survivorship Bias.
Look-Ahead Bias.
Time-Period Bias.
Data Mining Bias.
Sample data in investments can have a variety of problems. Survivorship bias occurs if companies are excluded from the analysis because they have gone out of business or because of reasons related to poor performance. Data-mining bias comes from finding models by repeatedly searching through databases for patterns. Look-ahead bias exists if the model uses data not available to market participants at the time the market participants act in the model. Finally, time-period bias is present if the time period used makes the results time-period specific or if the time period used includes a point of structural change.
You have developed a set of criteria for evaluating distressed credits. Companies that do not receive a passing score are classed as likely to go bankrupt within 12 months. You gathered the following information when validating the criteria:
◾Forty percent of the companies to which the test is administered will go bankrupt within 12 months: P(nonsurvivor) = 0.40.
◾Fifty-five percent of the companies to which the test is administered pass it: P(pass test) = 0.55.
◾The probability that a company will pass the test given that it will subsequently survive 12 months, is 0.85: P(pass test | survivor) = 0.85.
A.What is P(pass test | nonsurvivor)?
B.Using Bayes’ formula, calculate the probability that a company is a survivor, given that it passes the test; that is, calculate P(survivor | pass test).
C.What is the probability that a company is a nonsurvivor, given that it fails the test?
D.Is the test effective?
•We can set up the equation using the total probability rule:
P(passtest)=P(passtest∣∣survivor)P(survivor)+P(passtest∣∣nonsurvivor)P(nonsurvivor)
WeknowthatP(survivor)=1−P(nonsurvivor)=1−0.40=0.60.Therefore,P(passtest)=0.55=0.85(0.60)+P(passtest∣∣nonsurvivor)(0.40).ThusP(passtest∣∣nonsurvivor)=[0.55−0.85(0.60)]/0.40=0.10.
•P(survivor∣∣passtest)=[P(passtest∣∣survivor)/P(passtest)]P(survivor)=(0.85/0.55)0.60=0.927273
The information that a company passes the test causes you to update your probability that it is a survivor from 0.60 to approximately 0.927.
•According to Bayes’ formula, P(nonsurvivor | fail test) = [P(fail test | nonsurvivor)/ P(fail test)]P(nonsurvivor) = [P(fail test | nonsurvivor)/0.45]0.40.
We can set up the following equation to obtain P(fail test | nonsurvivor):
P(failtest)=P(failtest∣∣nonsurvivor)P(nonsurvivor)+P(failtest∣∣survivor)P(survivor)0.45=P(failtest∣∣nonsurvivor)0.40+0.15(0.60)
where P(fail test | survivor) = 1 − P(pass test | survivor) = 1 − 0.85 = 0.15. So P(fail test | nonsurvivor) = [0.45 − 0.15(0.60)]/0.40 = 0.90. Using this result with the formula above, we find P(nonsurvivor | fail test) = (0.90/0.45)0.40 = 0.80. Seeing that a company fails the test causes us to update the probability that it is a nonsurvivor from 0.40 to 0.80.
•A company passing the test greatly increases our confidence that it is a survivor. A company failing the test doubles the probability that it is a nonsurvivor. Therefore, the test appears to be useful.
na binomial como é que calculamos a média e o desvio padrão?
média : n*p
Variância: np(p-1)
A binomial segue uma função cumulativa nCr
Para normalizar uma variável X para uma distribuição normal (0,1)
Z=(X-média)/Desvpad
Multiplication rule
The number of ways to assign every member of a group of size n to n slots is …..
multinomial formula.
combination formula
permutation formula.
- The multiplication rule of counting says, for example, that if the first step in a process can be done in 10 ways, the second step, given the first, can be done in 5 ways, and the third step, given the first two, can be done in 7 ways, then the steps can be carried out in (10)(5)(7) = 350 ways.
- The number of ways to assign every member of a group of size n to n slots is n!
- The number of ways that n objects can be labeled with k different labels, with n1 of the first type, n2 of the second type, and so on, with n1 + n2 + … + nk = n, is given by n!/(n1!n2! … nk!). This expression is the multinomial formula.
- A special case of the multinomial formula is the combination formula. The number of ways to choose r objects from a total of n objects, when the order in which the r objects are listed does not matter, is
nCr= n!/(n−r)!r!
•The number of ways to choose r objects from a total of n objects, when the order in which the r objects are listed does matter, is This expression is the permutation formula.
nPr=n!/(n−r)!
Safety First Ratio
(E(Rp)- RL)/ DesvP
RL - limite mínimo em %
Label each of the following as an empirical, a priori, or subjective probability.
A.The probability that US stock returns exceed long-term corporate bond returns over a 10-year period, based on Ibbotson Associates data.
B.An updated (posterior) probability of an event arrived at using Bayes’ formula and the perceived prior probability of the event.
C.The probability of a particular outcome when exactly 12 equally likely possible outcomes exist.
D.A historical probability of default for double-B rated bonds, adjusted to reflect your perceptions of changes in the quality of double-B rated issuance.
A.The probability is an empirical probability.
B.The probability is a subjective probability.
C.The probability is an a priori probability.
D.The probability is a subjective probability.
Suppose X, Y, and Z are discrete random variables with these sets of possible outcomes: X = {2, 2.5, 3}, Y = {0, 1, 2, 3}, and Z = {10, 11, 12}. For each of the functions f(X), g(Y), and h(Z), state whether the function satisfies the conditions for a probability function.
A.f(2) = −0.01 f(2.5) = −0.50 f(3) = −0.51
B.g(0) = 0.25 g(1) = 0.50 g(2) = 0.125 g(3) = 0.125
C.h(10) = 0.35 h(11) = 0.15 h(12) = 0.52
•Because f(2) = −0.01 is negative, f(X) cannot be a probability function. Probabilities are numbers between 0 and 1.
•The function g(Y) does satisfy the conditions of a probability function:
All the values of g(Y) are between 0 and 1, and the values of g(Y) sum to 1.
•The function h(Z) cannot be a probability function: The values of h(Z) sum to 1.02, which is more than 1.
Over the last 10 years, a company’s annual earnings increased year over year seven times and decreased year over year three times. You decide to model the number of earnings increases for the next decade as a binomial random variable.
A.What is your estimate of the probability of success, defined as an increase in annual earnings?
For Parts B, C, and D of this problem, assume the estimated probability is the actual probability for the next decade.
B.What is the probability that earnings will increase in exactly 5 of the next 10 years?
C.Calculate the expected number of yearly earnings increases during the next 10 years.
D.Calculate the variance and standard deviation of the number of yearly earnings increases during the next 10 years.
E.The expression for the probability function of a binomial random variable depends on two major assumptions. In the context of this problem, what must you assume about annual earnings increases to apply the binomial distribution in Part B? What reservations might you have about the validity of these assumptions?
- The probability of an earnings increase (success) in a year is estimated as 7/10 = 0.70 or 70 percent, based on the record of the past 10 years.
- The probability that earnings will increase in 5 out of the next 10 years is about 10.3 percent. Define a binomial random variable X, counting the number of earnings increases over the next 10 years. From Part A, the probability of an earnings increase in a given year is p = 0.70 and the number of trials (years) is n = 10. Equation 1 gives the probability that a binomial random variable has x successes in n trials, with the probability of success on a trial equal to p.
P(X=x)=(nx)px(1−p)n−x=n!(n−x)!x!px(1−p)n−x
For this example,
(105)0.750.310−5=10!(10−5)!5!0.750.310−5=252×0.16807×0.00243=0.102919
We conclude that the probability that earnings will increase in exactly 5 of the next 10 years is 0.1029, or approximately 10.3 percent.
- The expected number of yearly increases is E(X) = np = 10 × 0.70 = 7.
- The variance of the number of yearly increases over the next 10 years is σ2 = np (1 − p) = 10 × 0.70 × 0.30 = 2.1. The standard deviation is 1.449 (the positive square root of 2.1).
- You must assume that 1) the probability of an earnings increase (success) is constant from year to year and 2) earnings increases are independent trials. If current and past earnings help forecast next year’s earnings, Assumption 2 is violated. If the company’s business is subject to economic or industry cycles, neither assumption is likely to hold.
A client has a portfolio of common stocks and fixed-income instruments with a current value of £1,350,000. She intends to liquidate £50,000 from the portfolio at the end of the year to purchase a partnership share in a business. Furthermore, the client would like to be able to withdraw the £50,000 without reducing the initial capital of £1,350,000. The following table shows four alternative asset allocations.
Mean and Standard Deviation for Four Allocations
(in Percent)
A B C D
Expected annual return 16 12 10 9
Standard deviation of return 24 17 12 11
Address the following questions (assume normality for Parts B and C):
A.Given the client’s desire not to invade the £1,350,000 principal, what is the shortfall level, RL? Use this shortfall level to answer Part B.
B.According to the safety-first criterion, which of the allocations is the best?
C.What is the probability that the return on the safety-first optimal portfolio will be less than the shortfall level, RL?
•Because £50,000/£1,350,000 is 3.7 percent, for any return less than 3.7 percent the client will need to invade principal if she takes out £50,000. So RL = 3.7 percent.
•To decide which of the allocations is safety-first optimal, select the alternative with the highest ratio [E(RP) − RL]/σP:
Allocation A: 0.5125 = (16 – 3.7)/24
Allocation B: 0.488235 = (12 – 3.7)/17
Allocation C: 0.525 = (10 – 3.7)/12
Allocation D: 0.481818 = (9 – 3.7)/11
Allocation C, with the largest ratio (0.525), is the best alternative according to the safety-first criterion.
•To answer this question, note that P(RC
At the end of the current year, an investor wants to make a donation of $20,000 to charity but does not want the year-end market value of her portfolio to fall below $600,000. If the shortfall level is equal to the risk-free rate of return and returns from all portfolios considered are normally distributed, will the portfolio that minimizes the probability of failing to achieve the investor’s objective most likely have the:
highest safety-first ratio? highest Sharpe ratio?
A No Yes
B Yes No
C Yes Yes
C is correct. The portfolio with the highest safety-first ratio minimizes the probability that the portfolio return will be less than the shortfall level (given normality). In this problem, the shortfall level is equal to the risk-free rate of return and thus the highest safety-first ratio portfolio will be the same as the highest Sharpe ratio portfolio.
An analyst stated that normal distributions are suitable for describing asset returns and that lognormal distributions are suitable for describing distributions of asset prices. The analyst’s statement is correct in regard to:
A.both normal distributions and lognormal distributions.
B.normal distributions, but incorrect in regard to lognormal distributions.
C.lognormal distributions, but incorrect in regard to normal distributions.
A is correct. A normal distribution is suitable for describing asset returns. However, the normal distribution is not suitable for asset prices because asset prices cannot be negative. The lognormal distribution is bounded by zero (skewed to the right) and is suitable for describing distributions of asset prices.
Peter Biggs wants to know how growth managers performed last year. Biggs assumes that the population cross-sectional standard deviation of growth manager returns is 6 percent and that the returns are independent across managers.
A.How large a random sample does Biggs need if he wants the standard deviation of the sample means to be 1 percent?
B.How large a random sample does Biggs need if he wants the standard deviation of the sample means to be 0.25 percent?
A.The standard deviation or standard error of the sample mean is σX⎯⎯⎯=σ/n√
. Substituting in the values for σX⎯⎯⎯
and σ
, we have 1% = 6%/n√
, or n√
= 6. Squaring this value, we get a random sample of n = 36.
B.As in Part A, the standard deviation of sample mean is σX⎯⎯⎯=σ/n√
. Substituting in the values for σX⎯⎯⎯
and σ
, we have 0.25% = 6%/n√
, or n√
= 24. Squaring this value, we get a random sample of n = 576, which is substantially larger than for Part A of this question.
Petra Munzi wants to know how value managers performed last year. Munzi estimates that the population cross-sectional standard deviation of value manager returns is 4 percent and assumes that the returns are independent across managers.
A.Munzi wants to build a 95 percent confidence interval for the mean return. How large a random sample does Munzi need if she wants the 95 percent confidence interval to have a total width of 1 percent?
B.Munzi expects a cost of about $10 to collect each observation. If she has a $1,000 budget, will she be able to construct the confidence interval she wants?
•Assume the sample size will be large and thus the 95 percent confidence interval for the mean of a sample of manager returns is X⎯⎯⎯±1.96sX⎯⎯⎯
, where sX⎯⎯⎯=s/n√
. Munzi wants the distance between the upper limit and lower limit in the confidence interval to be 1 percent, which is
(média+1.96stdX)−(média−1.96stdX)=1%
Simplifying this equation, we get 2(1.96stdX)
= 1%. Finally, we have 3.92stdX
= 1%, which gives us the standard deviation of the sample mean, stdX
= 0.255%. The distribution of sample means is stdX=s/n√
. Substituting in the values for stdX
and s, we have 0.255% = 4%/n√
, or n√
= 15.69. Squaring this value, we get a random sample of n = 246.
•With her budget, Munzi can pay for a sample of up to 100 observations, which is far short of the 246 observations needed. Munzi can either proceed with her current budget and settle for a wider confidence interval or she can raise her budget (to around $2,460) to get the sample size for a 1 percent width in her confidence interval.
Assume that the equity risk premium is normally distributed with a population mean of 6 percent and a population standard deviation of 18 percent. Over the last four years, equity returns (relative to the risk-free rate) have averaged −2.0 percent. You have a large client who is very upset and claims that results this poor should never occur. Evaluate your client’s concerns.
A.Construct a 95 percent confidence interval around the population mean for a sample of four-year returns.
B.What is the probability of a −2.0 percent or lower average return over a four-year period?
•This is a small-sample problem in which the sample comes from a normal population with a known standard deviation; thus we use the z-distribution in the solution. For a 95 percent confidence interval (and 2.5 percent in each tail), the critical z-value is 1.96. For returns that are normally distributed, a 95 percent confidence interval is of the form
μ+1.96σn√
The lower limit is X1=μ−1.96σn√=6%−1.9618%4√
= 6% – 1.96(9%) = –11.64%.
The upper limit is Xu=μ+1.96σn√=6%+1.9618%4√
= 6% + 1.96(9%) = 23.64%.
There is a 95 percent probability that four-year average returns will be between −11.64 percent and +23.64 percent.
•The critical z-value associated with the −2.0 percent return is
Z=X⎯⎯⎯−μσ/n√=−2%−6%18%/4√=−8%9%=−0.89
Using a normal table, the probability of a z-value less than −0.89 is P(Z
Compare the standard normal distribution and Student’s t-distribution.
(Refer to Figure 1 to help visualize the answer to this question.) Basically, only one standard normal distribution exists, but many t-distributions exist—one for every different number of degrees of freedom. The normal distribution and the t-distribution for a large number of degrees of freedom are practically the same. The lower the degrees of freedom, the flatter the t-distribution becomes. The t-distribution has less mass (lower probabilities) in the center of the distribution and more mass (higher probabilities) out in both tails. Therefore, the confidence intervals based on t-values will be wider than those based on the normal distribution. Stated differently, the probability of being within a given number of standard deviations (such as within ±1 standard deviation or ±2 standard deviations) is lower for the t-distribution than for the normal distribution.
An exchange rate has a given expected future value and standard deviation.
A. Assuming that the exchange rate is normally distributed, what are the probabilities that the exchange rate will be at least 2 or 3 standard deviations away from its mean?
B. Assume that you do not know the distribution of exchange rates. Use Chebyshev’s inequality (that at least 1 − 1/k2 proportion of the observations will be within k standard deviations of the mean for any positive integer k greater than 1) to calculate the maximum probabilities that the exchange rate will be at least 2 or 3 standard deviations away from its mean.
A.The probabilities can be taken from a normal table, in which the critical z-values are 2.00 or 3.00 and we are including the probabilities in both tails. The probabilities that the exchange rate will be at least 2 or 3 standard deviations away from the mean are
P( ∣∣ X−μ ∣∣≥2σ)=0.0456P( ∣∣ X−μ ∣∣≥3σ)=0.0026
B.With Chebyshev’s inequality, the maximum probability of the exchange rate being at least k standard deviations from the mean is P(| X − μ | ≥ kσ) ≤ (1/k)2. The maximum probabilities of the rate being at least 2 or 3 standard deviations away from the mean are
P( ∣∣ X−μ ∣∣≥2σ)≤(1/2)2=0.2500P( ∣∣ X−μ ∣∣≥3σ)≤(1/3)2=0.1111
The probability of the rate being outside 2 or 3 standard deviations of the mean is much smaller with a known normal distribution than when the distribution is unknown and we are relying on Chebyshev’s inequality.
Suppose we take a random sample of 30 companies in an industry with 200 companies. We calculate the sample mean of the ratio of cash flow to total debt for the prior year. We find that this ratio is 23 percent. Subsequently, we learn that the population cash flow to total debt ratio (taking account of all 200 companies) is 26 percent. What is the explanation for the discrepancy between the sample mean of 23 percent and the population mean of 26 percent?
A.Sampling error.
B.Bias.
C.A lack of consistency.
A is correct. The discrepancy arises from sampling error. Sampling error exists whenever one fails to observe every element of the population, because a sample statistic can vary from sample to sample. As stated in the reading, the sample mean is an unbiased estimator, a consistent estimator, and an efficient estimator of the population mean. Although the sample mean is an unbiased estimator of the population mean—the expected value of the sample mean equals the population mean—because of sampling error, we do not expect the sample mean to exactly equal the population mean in any one sample we may take.
Hand Associates manages two portfolios that are meant to closely track the returns of two stock indexes. One index is a value-weighted index of 500 stocks in which the weight for each stock depends on the stock’s total market value. The other index is an equal-weighted index of 500 stocks in which the weight for each stock is 1/500. Hand Associates invests in only about 50 to 100 stocks in each portfolio in order to control transactions costs. Should Hand use simple random sampling or stratified random sampling to choose the stocks in each portfolio?
Hand Associates should use stratified random sampling for its portfolio that tracks the value-weighted index. Using 50–100 stocks to track 500 means that Hand will invest in all or almost all of the largest stocks in the index and few of the smallest. In addition to size, the stocks may be grouped by industry, riskiness, and other traits, and Hand Associates may select stocks to represent each of these groups or strata. For the equal-weighted index, Hand can use simple random sampling, in which each stock is equally likely to be chosen. Even in this case, however, Hand could use stratified random sampling to make sure it is choosing stocks that represent the various factors underlying stock performance.
There are four possible outcomes when we test a null hypothesis:
- We reject a false null hypothesis. This is a correct decision.
- We reject a true null hypothesis. This is called a Type I error.
- We do not reject a false null hypothesis. This is called a Type II error.
- We do not reject a true null hypothesis. This is a correct decision.
These are mutually exclusive errors: If we mistakenly reject the null, we can only be making a Type I error; if we mistakenly fail to reject the null, we can only be making a Type II error.
Test Statistic for Tests Concerning the Value of a Population Variance (Normal Population). If we have n independent observations from a normally distributed population, the appropriate test statistic is
X^2=((n-1)*s^2)/σ0^2
You continue with your analysis of Sendar Equity Fund, a midcap growth fund that has been in existence for only 24 months. Recall that during this period, Sendar Equity achieved a sample standard deviation of monthly returns of 3.60 percent. You now want to test a claim that the particular investment disciplines followed by Sendar result in a standard deviation of monthly returns of less than 4 percent.
- Formulate null and alternative hypotheses consistent with the verbal description of the research goal.
- Identify the test statistic for conducting a test of the hypotheses in Part 1.
- Identify the rejection point or points for the hypothesis tested in Part 1 at the 0.05 level of significance.
- Determine whether the null hypothesis is rejected or not rejected at the 0.05 level of significance. (Use the tables in the back of this volume.)
Solution to 1:
We have a “less than” alternative hypothesis, where σ is the underlying standard deviation of return on Sendar Equity Fund. Being careful to square standard deviation to obtain a test in terms of variance, the hypotheses are H0: σ2 ≥ 16.0 versus Ha: σ2
F test
Test Statistic for Tests Concerning Differences between the Variances of Two Populations (Normally Distributed Populations). Suppose we have two samples, the first with n1 observations and sample variance s1^2
, the second with n2 observations and sample variance s2^2
. The samples are random, independent of each other, and generated by normally distributed populations. A test concerning differences between the variances of the two populations is based on the ratio of sample variances
Equation (16)
F=s1^2/s2^2
Thus, if we conduct a two-sided test at the α = 0.01 level of significance, we need to find the rejection point in F-tables at the α/2 = 0.01/2 = 0.005 significance level for a one-sided test (Case 1). But a one-sided test at 0.01 uses rejection points in F-tables for α = 0.01 (Case 2). As an example, suppose we are conducting a two-sided test at the 0.05 significance level. We calculate a value of F of 2.77 with 12 numerator and 19 denominator degrees of freedom. Using the F-tables for 0.05/2 = 0.025 in the back of the volume, we find that the rejection point is 2.72. Because the value 2.77 is greater than 2.72, we reject the null hypothesis at the 0.05 significance level.
Suppose our chosen level of significance is 0.05 for a two-tailed test and we have a value of F of 0.11, with 7 numerator degrees of freedom and 9 denominator degrees of freedom. We take the reciprocal, 1/0.11 = 9.09. Then we look up this value in the F-tables for 0.025 (because it is a two-tailed test) with degrees of freedom reversed: F for 9 numerator and 7 denominator degrees of freedom. In other words, F9,7 = 1/F7,9 and 9.09 exceeds the critical value of 4.82, so F7,9 = 0.11 is significant at the 0.05 level.
You are investigating whether the population variance of returns on the KOSPI Index of the South Korean stock market changed subsequent to the global financial crisis that peaked in 2008. For this investigation, you are considering 2004 to 2006 as the pre-crisis period and 2010 to 2012 as the post-crisis period. You gather the data in Table 7 for 156 weeks of returns during 2004 to 2006 and 156 weeks of returns during 2010 to 2012. You have specified a 0.01 level of significance.
n Mean Week return(%) Mean Variance Before 156 0,358 7,240 After 156 0,11 6,269
- Formulate null and alternative hypotheses consistent with the verbal description of the research goal.
- Identify the test statistic for conducting a test of the hypotheses in Part 1.
- Determine whether or not to reject the null hypothesis at the 0.01 level of significance. (Use the F-tables in the back of this volume.)
We have a “not equal to” alternative hypothesis:
H0 :σ2Before=σ2After versus Ha :σ2Before≠σ2After
Solution to 2:
To test a null hypothesis of the equality of two variances, we use F=s21/s22
with 156 − 1 = 155 numerator and denominator degrees of freedom.
Solution to 3:
The “before” sample variance is larger, so following a convention for calculating F-statistics, the “before” sample variance goes in the numerator: F = 7.240/6.269 = 1.155. Because this is a two-tailed test, we use F-tables for the 0.005 level (= 0.01/2) to give a 0.01 significance level. In the tables in the back of the volume, the closest value to 155 degrees of freedom is 120 degrees of freedom. At the 0.01 level, the rejection point is 1.61. Because 1.155 is less than the critical value 1.61, we cannot reject the null hypothesis that the population variance of returns is the same in the pre- and post-global financial crisis periods.
Parametric and Non Parametric tets
- A parametric test is a hypothesis test concerning a parameter or a hypothesis test based on specific distributional assumptions. In contrast, a nonparametric test either is not concerned with a parameter or makes minimal assumptions about the population from which the sample comes.
- A nonparametric test is primarily used in three situations: when data do not meet distributional assumptions, when data are given in ranks, or when the hypothesis we are addressing does not concern a parameter.
Suppose we are testing a null hypothesis, H0, versus an alternative hypothesis, Ha, and the p-value for the test statistic is 0.031. At which of the following levels of significance—α = 0.10, α = 0.05, and/or α = 0.01—would we reject the null hypothesis?
By the definition of p-value, 0.031 is the smallest level of significance at which we can reject the null hypothesis. Because 0.031 is smaller than 0.10 and 0.05, we can reject the null hypothesis at the 0.10 and 0.05 significance levels. Because 0.031 is larger than 0.01, however, we cannot reject the null hypothesis at the 0.01 significance level.
Identify the theoretically correct test statistic to use for a hypothesis test concerning the mean of a single population under the following conditions:
A.The sample comes from a normally distributed population with known variance.
B.The sample comes from a normally distributed population with unknown variance.
C.The sample comes from a population following a non-normal distribution with unknown variance. The sample size is large.
A.When sampling from a normally distributed population with known variance, the correct test statistic for
hypothesis tests concerning the mean is the z-statistic.
B.When sampling from a normally distributed population with unknown variance, the theoretically correct test statistic for hypothesis tests concerning the mean is the t-statistic.
C.When the sample size is large, the central limit theorem applies. Consequently, the sample mean will be approximately normally distributed. When the population variance is not known, a test using the t-statistic is theoretically preferred. A test using the z-statistic is also sufficient when the sample size is large, as in this case.
All else equal, is specifying a smaller significance level in a hypothesis test likely to increase the probability of a:
Type I error? Type II error? A No No B No Yes C Yes No
•B is correct. Specifying a smaller significance level decreases the probability of a Type I error (rejecting a true null hypothesis), but increases the probability of a Type II error (not rejecting a false null hypothesis). As the level of significance decreases, the null hypothesis is less frequently rejected.
All else equal, is increasing the sample size for a hypothesis test likely to decrease the probability of a:
Type I error? Type II error?
A No Yes
B Yes No
C Yes Yes
•C is correct. The only way to avoid the trade-off between the two types of errors is to increase the sample size; increasing sample size (all else equal) reduces the probability of both types of errors. From the reading on sampling and estimations, all else equal, a larger sample size will decrease both the standard error and the width of the confidence interval. In other words, the precision of the estimate of the population parameter is increased.
