CFAI 2-Quantitative Methods Flashcards

Question

Intervalos de confiança padrão, 1 desvio padrão, 2 desvios padrões e 3 desvios padrões

Answer 1

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%.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

FV= A*([(1+r)^n-1]/r) PV = A*((1-x)/r) x= 1/(1+r)^n

Answer 6

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.

Answer 7

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).

Answer 8

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.

Answer 9

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

Answer 10

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.

Answer 11

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.

Answer 12

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.

Answer 13

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.

Answer 14

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).

Answer 15

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.

Answer 16

Use your calculator’s financial functions to verify that the stated interest rate of the savings account is 6 percent with monthly compounding.

Answer 17

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.)

Answer 18

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.

Answer 19

EPS grew at an annual rate of 15.02 percent during the four years.

Answer 20

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%.

Answer 21

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.

Answer 22

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.

Answer 23

* 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).

Answer 24

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 ```

Answer 25

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

Answer 26

* 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.

Answer 27

•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

Answer 28

•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.

Answer 29

•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).

Answer 30

•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.

Answer 31

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.

Answer 32

Retorno esperado= Pdefault*0+(1+r)*(1-Pdefault) Sendo que r é a taxa que a obrigação oferece.

Answer 33

[N*(N-1)]/2

Answer 34

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)](1/3) = 0.75 or 75 percent. This probability is identical to your estimate of P(DriveMed expands | EPS exceeded consensus).

Answer 35

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.

Answer 36

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.

Answer 37

n!/(n1! n2!…nk!) É parecida à nPr contudo dá label a todas. n! factorial dá-nos todas as opções.

Answer 38

* 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.

Answer 39

* 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.

Answer 40

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.

Answer 41

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.

Answer 42

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.

Answer 43

n*(n-1)/2 = 10

Answer 44

•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!).

Answer 45

nCr . Here,n=10 and r=4,so the answer is 10!/[(10−4)!4!]=3,628,800/(720)(24)= 210.

Answer 46

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. ```

Answer 47

•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.

Answer 48

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.

Answer 49

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.

Answer 50

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.

Answer 51

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.

Answer 52

•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.

Answer 53

média : n*p Variância: n*p*(p-1) A binomial segue uma função cumulativa nCr

Answer 54

Z=(X-média)/Desvpad

Answer 55

* 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)!

Answer 56

(E(Rp)- RL)/ DesvP RL - limite mínimo em %

Answer 57

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.

Answer 58

•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.

Answer 59

* 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.

Answer 60

•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

Answer 61

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.

Answer 62

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.

Answer 63

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.

Answer 64

•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.

Answer 65

•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

Answer 66

(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.

Answer 67

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.

Answer 68

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.

Answer 69

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.

Answer 70

* 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.

Answer 71

X^2=((n-1)*s^2)/σ0^2

Answer 72

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

Answer 73

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.

Answer 74

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.

Answer 75

* 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.

Answer 76

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.

Answer 77

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.

Answer 78

•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.

Answer 79

•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.