SOA Probability

Question 1

If A ⊂ B then (A n B)

Answer 1

(A n B) = A

Question 2

Probability Generating Function Defined as PGF where

P_x(t) =

Answer 2

P_x(t) = E [t^X]

Question 3

E ( X ( X -1 ) ) = 2nd Moment of what Generating Function?

Answer 3

PGF - Probability Generating Function

Question 4

E [X ∣ j ≤ X ≤ k] - continuous case

Answer 4

Integrate numerator from j to k

( ∫ x ⋅ f_X (x) dx )

÷
( Pr ( j ≤ X ≤ k ) )

Question 5

Percentile for Discrete Random Variables

Answer 5

F_x(π_p) ≥ p

i.e the function at π_p has to atleast be equal or greater than the percentile p

Question 6

E [X | j ≤ X ≤ k] - Discrete Case

Answer 6

Sum numerator from x = j to k

( ∑ (x)( Pr [j ≤ X ≤ k] )

÷

( Pr [j ≤ X ≤ k] )

Question 7

Percentile for Continous Random Variable

Answer 7

density function f_X(π_p) = p

Has to equal the percentile

Question 8

Finding mode of discrete random variable

Answer 8

calculate probabilities of each possible value and choose the one that gives the largest probability

Question 9

Finding mode of continuous random variable

Answer 9

take derivative of density function set it equal to 0 and solve for mode. (finding local maximum of function)

Question 10

Cumulative Distribution Function (CDF) of a probability density function (PDF)

Answer 10

integrate from lowest value of X to the variable x itself

0 <f></f>

∫ f(t) dt

Question 11

Chebyshev’s Inequality

Answer 11

Pr( |X-µ| ≥ kσ ) ≤ ( 1 / k² )

Question 12

How to break up the inequality of Chebyshev’s Equation

Answer 12

Pr( |X-µ| ≥ kσ )

=

Pr( (X-µ) ≥ kσ ) + Pr( (X-µ) ≤ -kσ )

Question 13

Univariate Transformation CDF METHOD

From X to Y

Answer 13

1.) Given PDF of X find CDF of X

2.) Perform Transformation where F_Y( y ) = P( Y ≤ y ) with subsitution

3.) Restate CDF of Y using CDF of X ,

then subsitute CDF of X found in step 1 into CFD of Y

4.) Take Derivative of CDF of Y to find PDF of Y

Question 14

Univariate Transformation PDF METHOD

From X to Y

Answer 14

1.) Get PDF of X if not given

2.) Find PDF of Y using the formula

f_Y( y ) = f_X( [g^-1( y )] ) • | (d/dy) g^-1( y ) |

3.) Integrate PDF of Y to get CDF of Y if required

Question 15

Discrete Uniform PMF

Answer 15

( 1 / b - a + 1)

Question 16

Discrete Uniform E[X]

Answer 16

( a + b / 2 )

Question 17

Discrete Uniform Var[X]

Answer 17

[( b - a + 1 )² - 1]

÷

12

Question 18

Bernoulli’s E[X]

Answer 18

p

Question 19

Bernoulli’s Var[X]

Answer 19

pq

Question 20

Bernoulli’s MGF

Answer 20

pe^t + q

Question 21

Bernoull’s Variance Short-cut for Y = (a-b)X + b

Answer 21

(b - a)²• pq

Question 22

Property of Expected One RV: E[c]=

Answer 22

E[c]=c, c = constant

Question 23

Property of Expected One RV: E[c⋅g(X)]=

Answer 23

E[c⋅g(X)]= c ⋅ E[g(X)]

Question 24

Property of Expected One RV: E[g₁(X)+g₂(X)+…+g_k(X)] =

Answer 24

E[g₁(X)+g₂(X)+…+g_k(X)]

=

E[g₁(X)] + E[g₂(X)]+ …+E[g_k(X)]

Question 25

Variance formula for One RV:

Var[X]

Var[g(X)]

Answer 25

Var[X] = E[(X−μ)²] = E[X²] − (E[X])²

Var[g(X)] = E[(g(X) − E[g(X)])²] = E[g(X)²] − (E[g(X)])²​

Question 26

Property of Variance One RV: Var[c] =

Answer 26

Var[c] = 0,

c = constant

Question 27

Property of Variance One RV: Var [aX + b]

Answer 27

Var [aX + b] =

a² • Var[X]

a,b = constant

Question 28

Coefficient of Variation for One RV

CV[X]=

Answer 28

CV[X] =

(SD[X])

• *÷**
• *(E[X])**

Question 29

Binomial Mean E[X]

Answer 29

E[X] = np

Question 30

Binomial Variance Var[X]

Answer 30

Var[X] = npq

Question 31

Binomial MGF

Answer 31

M_X(t) = (pe^t+q)ⁿ

Question 32

Binomial PGF

Answer 32

P_X(t) = (p^t+q)ⁿ

Question 33

Hypergeometric PMF: Pr(X = x)

Answer 33

Pr(X = x)

[( m Choose x ) • ( N - m Choose n - x )]

÷​

( N choose n )

x = sucess out of m = total sucess

n - x = failures out of N - m = total failures

N = population size; n = sample size

Question 34

Geometric PMF

Answer 34

Pr(X = x) = (1−p)^x−1 • p

Where the first success is observed

Define X as the number of rolls to get

Question 35

Geometic E[X]

Answer 35

E[X] = ( 1 / p )

Question 36

Geometric Var[X]

Answer 36

Var[X] = [(1 - p)÷p²]

Question 37

Geometric MGF

Answer 37

M_X(t) =

(pe^t)

÷

(1−(1−p)e^t )

for t < −ln(1−p)

Question 38

Memoryless Property of a Distribution

Answer 38

The memoryless property states that, for positive integer c,

Pr( X − c = x∣ X > c ) =

Pr(X = x )​

Question 39

Negative Binomial PMF

Answer 39

Pr(X = x) =

(x-1 choose r−1) •p^r•(1−p)^x−r

r = the desired number of “successes”

p = probability of a success

X = number of “trials” until the r^th “success”

X represent the number of coin tosses necessary for three heads to occur

Question 40

Negative Binomial E[X]

Answer 40

E[X] = r ÷ p

Question 41

Negative Binomial Var[X]

Answer 41

Var[X] = r•( 1−p ÷ p²)

Question 42

Negative Binomial MGF

Answer 42

MX(t) =

(pe^t ÷ 1 − (1 − p)e^t)^r

Question 43

Geometric Distribution for Pr( X ≥ x )

Answer 43

∑ (1-p)^x-1 • p

=

(1-p)^x-1

Question 44

Geometric Distribution Derivation

Answer 44

Independent Bernoulli trials​

P(X=x)

= Pr( first “success” on x^th “trial”)

= Pr (“failure” on the first x-1 “trials” and “success” on x^th “trial”)

= Pr( “failure” on the first x-1 “trials”) • Pr( “success” on the x^th “trial”)

= (1 - p)^x-1 • p

Question 45

Geometric Alternative Form (failures)

Answer 45

Let Y be the number of “failures” before the first “success”, rather than the number of “trials”

number of “trials” = number of “failures’’ + number of “successes”

X = Y+ 1 ⇒ Y = X − 1

Let Y be the number of rolls before getting...

Question 46

Negative Binomial Distribution Derivation

Answer 46

Independent Bernoulli Trials

Pr (X = x )

= Pr (r^th “success’’ on x^th “trial”)

= Pr( r−1 “successes’’ on the first x−1 “trials” ∩ “success’’ on x^th “trial” )​

= Pr(r−1 “successes’’ on the first x−1 “trials” )•Pr(“success’’ on x^th “trial”)

Question 47

Negative Binomial Alternative Form “failures”

Answer 47

Let Y be the number of “failures” before the r^th “success”

= Pr(X − r = y)

= Pr(X = y + r)

Let Y represent the number of tails before getting the third head

Question 48

Exponential MGF

Answer 48

M_x(t) = ( 1 / 1−θt )

t < (1 / θ)

Question 49

Exponential Var[X]

Answer 49

X∼Exponential(θ) = θ²

X∼Exponential(λ) = (1 / λ²)

Question 50

X∼Exponential(θ) E[X]

Answer 50

E[X]; X∼Exponential(θ) = θ

E[X]; X∼Exponential(λ) = λ

Question 51

Gamma PDF

Answer 51

f_X(x) =

(1 / Γ(α)) ⋅ (x^α−1/θ^α) ⋅ e^{(−x / θ)}

if α is a positive integer then

Γ(α) = (α - 1)!

Question 52

Gamma E[X]

Answer 52

E[X] = αθ

Question 53

Gamma Var[X]

Answer 53

Var[X] = αθ²

Question 54

Gamma MGF

Answer 54

M_X(t) = (1 / 1−θt)^α

Question 55

Normal Distribution PDF

Answer 55

f_X(x)=

[(1 / (σ • sqrt(2π))] •

( e^{−[((x−μ)^2) / (2 • σ^2)]} )

Question 56

Standard Normal Distribution

Answer 56

f_Z(z) =

(1 / sqrt(2π) •

(e^{−((z)^2) / 2 )} )

Z = ( X - μ ) / ( σ )

Question 57

Joint Density Function for Pr( X + Y < 2 )

inner integral limit

Answer 57

double integration:

Determine limits for inner integral following the picture

∫ ( ∫ f_X,Y(x,y) dy ) dx

Question 58

Pr(X ≤ c ∣ Y = y)

Answer 58

Integrate from −∞ to c

∫ f_X∣Y(x∣y) dx

=

∫ [fX,Y(x,y) / fY(y)] dx

Question 59

Pr(X ≤ x ∣ Y ≤ y)

Answer 59

[ Pr(X ≤ x _∩ Y ≤ y)

/ Pr(Y ≤ y) ]

Question 60

Weighted Average of CDF

Answer 60

F_Y(y) = a₁F_C1(y) + a₂F_C2(y)

Question 61

Weighted Average of Survival Function

Answer 61

S_Y(y) = a₁S_C1(y) + a₂S_C2(y)

Question 62

To construct the mixed (or unconditional) distribution of Y

Answer 62

Pr(Y = y)

=

Pr(Y = y∣X = x)⋅Pr(X = x) + Pr(Y = y∣X = x)⋅Pr(X = x)

Question 63

Pr(A ∣ B) + Pr(A′ ∣ B) =

Answer 63

[ Pr(A_∩B) + Pr(A′_∩B)

/

Pr(B) ]

Question 64

Pr(A ∣ B) + Pr(A′ ∣ B) =

Answer 64

[Pr(B)

/

Pr(B) ]

=

1

Question 65

Double Expectation

Answer 65

E[X] = E[E [X ∣ Y] ]

Question 66

Law of total Variance

Answer 66

Var[X]

=

E[Var[X | Y]] + Var[E[X ∣ Y]]

EVVE

Question 67

Pr(X = x∣Y ≤ y) =

Answer 67

Pr(X=x _∩ Y≤y)

/

Pr(Y ≤ y)

Question 68

Pr(X = x∣Y = y)

Answer 68

Pr(X=x _∩Y=y)

/

Pr(Y=y)

Question 69

Cov[X,X]

Answer 69

E[X⋅X] − E[X]⋅E[X]

=

E[X2] − (E[X])²

=

Var[X]

Question 70

Cov[a,X]

Answer 70

0

Question 71

Cov[a,b]

Answer 71

0

Question 72

Cov[aX , bY]

Answer 72

ab⋅Cov[X , Y]

Question 73

Var[aX]

Answer 73

Cov[aX,aX]

=

a²⋅Cov[X,X]

=

a²⋅Var[X]

Question 74

Cov[X+a , Y+b]

Answer 74

Cov[X , Y]

Question 75

Cov[aX + bY , cP + dQ]

Answer 75

ac⋅Cov[X,P] + ad⋅Cov[X,Q] + bc⋅Cov[Y,P] + bd⋅Cov[Y,Q]

Question 76

Var[aX + bY]

Answer 76

a²⋅Var[X] + 2ab⋅Cov[X,Y] + b²⋅Var[Y]

Question 77

Cov[X,Y] =

Answer 77

E[XY] − E[X]⋅E[Y]

Question 78

ρX,Y=Corr[X,Y]

Coefficient of Correlation

Answer 78

Cov[X,Y]

/

SD[X]⋅SD[Y]

Question 79

Multivariate Transformation CDF Method

Case 1: Transforming two variables (X and Y) to one variable (W).

Answer 79

1.) Using equation of transformation, W=g(X,Y), express F_W(w)=Pr(W≤w) = Pr[g(X,Y)≤w]

.

2. ) Calculate F_W(w)=Pr(W≤w) by integrating over region of integration defined by domain of f_X,_Y(x,y) and g(X,Y)≤w
   (transformation) .
3. ) Differentiate F_W(w)

to get f_W(w) if required.

Question 80

Multivariate Transformation PDF Method

Case 2: Transforming two variables (X1 and X2) to two variables (W1 and W2)

Answer 80

1.) Find f_X1,_X2(x1,x2)

if not given.

2. ) Introduce dummy variable (for Case 1 only).
3. ) Find inverse of equations of transformation, h1(w1,w2)

and h2(w1,w2)

4.)Calculate determinant of Jacobian matrix, J and take its absolute value

5.) Find f_W1,_W2(w1,w2) using the following formula.
f_W1,_W2(w1,w2) = f_X1,_X2[h1(w1,w2),h2(w1,w2)]⋅|J|, J≠0

Question 81

Combination Formula

Answer 81

ₙCₖ = [n! / (n-k)! * k!]

Question 82

Permutation Formula

Answer 82

ₙPₖ = [n! / (n-k)!]

Question 83

Combinatorial Probability

Answer 83

Probability =(Number of Permutations or Combination Satisfying requirements) / (Total Number of Permutations or Combinations)