Probability Flashcards

Question

Prove that A and B' are independent given A and B are independent

Answer 1

We have A = (A ∩ B) ∪ (A ∩ B'), where A ∩ B and A ∩ B' are disjoint, so using the independence of A and B, P(A ∩ B') = P (A) − P(A ∩ B) = P(A) − P(A) P(B) = P (A) (1 − P(B)) = P(A)P(B')

Answer 2

if 1. Ω = ∪ᵢ≥₁ Bᵢ (so that at least one Bi must happen), and 2. Bᵢ ∩ Bⱼ = ∅ whenever i ≠ j (so that no two can happen together)

Answer 3

Suppose {B1, B2, . . .} is a partition of Ω by sets from F, such that P (Bᵢ) > 0 for all i ≥ 1. Then for any A ∈ F P(A) = ᵢ≥₁ΣP(A|Bᵢ)P(Bᵢ)

Answer 4

P(A) = P(A ∩ (∪ᵢ≥₁Bᵢ)), since ∪ᵢ≥₁Bᵢ = Ω = P(∪ᵢ≥₁(A ∩ Bᵢ)) = ᵢ≥₁Σ P (A ∩ Bᵢ) by axiom P3, since A ∩ Bᵢ, i ≥ 1 are disjoint = ᵢ≥₁Σ P (A|Bᵢ)P(Bᵢ)

Answer 5

Suppose that {B1, B2, . . .} is a partition of Ω by sets from F such that P (Bi) > 0 for all i ≥ 1. Then for any A ∈ F such that P (A) > 0 P(Bₖ|A) = P(A|Bₖ)P(Bₖ)/(ᵢ≥₁Σ P (A|Bᵢ)P(Bᵢ))

Answer 6

We have P(Bₖ|A) = P(Bₖ ∩ A)/P(A) = P(A|Bₖ)P(Bₖ)/P(A) Now substitute for P(A) using the law of total probability

Answer 7

Answer 8

P(A ∩ B) = P(A|B) P(B) = P(B|A) P(A)

Answer 9

P (A1 ∩ A2 ∩ . . . ∩ An) = = P(A1) P(A2|A1). . . P(An|A1 ∩ A2 ∩ . . . ∩ An−1)

Answer 10

P (A1 ∪ A2 ∪ . . . ∪ An) = ⁿΣᵢ₌₁ P(Aᵢ) - Σᵢ>ⱼ P(Ai ∩ Aj) + ... + (-1)ⁿ⁺¹P(A1 ∩ A2 ∩ . . . ∩ An)

Answer 11

``` A discrete random variable X on a probability space (Ω, F, P) is a function X : Ω → R such that (a) {ω ∈ Ω : X(ω) = x} ∈ F for each x ∈ R, (b) ImX := {X(ω) : ω ∈ Ω} is a finite or countable subset of R ```

Answer 12

The probability mass function (p.m.f.) of X is the function pₓ : R → [0, 1] defined by pₓ(x) = P(X = x)

Answer 13

If x ≠ ImX X (that is, X(ω) never equals x) then pₓ(x) = P ({ω : X(ω) = x}) = P (∅) = 0.

Answer 14

ₓ∈ᵢₘₓΣ pₓ(x) = ₓ∈ᵢₘₓΣ P ({ω : X(ω) = x}) =P(ₓ∈ᵢₘₓ ∪ {ω : X(ω) = x}) since the events are disjoint = P (Ω) since every ω ∈ Ω gets mapped somewhere in ImX = 1

Answer 15

P(X = 0) = 1 − p, P(X = 1) = p

Answer 16

P (X = k) = nCk p^k (1-p)^n-k

Answer 17

X ∼ Ber(p)

Answer 18

X ∼ Bin(n, p)

Answer 19

X ∼ Geom(p)

Answer 20

X ∼ Po(λ)

Answer 21

``` P(X = k) = p(1 − p)^k-1, k = 1, 2, .... ```

Answer 22

We can use X to model the number of independent trials needed until we see the first success, where p is the probability of success on a single trial

Answer 23

P (Y = k) = p(1 − p)^k, | k = 0, 1, ...

Answer 24

P (X = k) = ( λ^k e^-λ) /k!, k = 0, 1, ...

Answer 25

``` The expectation (or expected value or mean) of X is E[X] = ₓ∈ᵢₘₓΣ xP(X=x) provided that ₓ∈ᵢₘₓΣ |x|P(X=x) < ∞ ```

Answer 26

E [h(X)] = ₓ∈ᵢₘₓΣ h(x)P (X = x) | provided that ₓ∈ᵢₘₓΣ |h(x)|P (X = x) < ∞.

Answer 27

Let A = {y : y = h(x) for some x ∈ ImX} Start from the rhs. Write it as two sums, one over y∈A, the other over x∈ImX:h(x)=y pg22

Answer 28

The kth moment of X, when it exists

Answer 29

If X is non-negative then E [X] ≥ 0 We have ImX ⊆ [0, ∞) and so E [X] = ₓ∈ᵢₘₓΣ xP (X = x) is a sum whose terms are all non-negative and so must itself be non-negative.

Answer 30

E [aX + b] = aE [X] + b

Answer 31

``` For a discrete random variable X, the variance of X is defined by var (X) = E[(X − E[X])² ] = E[X²] - (E[X])² provided that this quantity exists. ```

Answer 32

The variance is a measure of how much the distribution of X is spread out about its mean: the more the distribution is spread out, the larger the variance.

Answer 33

Yes since (X−E [X])2 is a non-negative random variable, var (X) ≥ 0

Answer 34

Standard deviation^2 = var (X)

Answer 35

var (Y ) = var (aX + b) = a² var (X)

Answer 36

P(X = x|B) = P({X = x} ∩ B) / P(B), for x ∈ R

Answer 37

ₓΣxP(X = x|B), whenever the sum converges absolutely We write pₓ|ᵦ(x) = P(X=x|B)

Answer 38

If {B1, B2, . . .} is a partition of Ω such that P (Bi) > 0 for all i ≥ 1 then E [X] = ᵢ≥₁ΣE [X | Bᵢ] P(Bᵢ), whenever E [X] exists.

Answer 39

Use the total law of probability to split into two sums, one over x, one over i. pg24

Answer 40

pₓ,ᵧ (x, y) = P ({X = x} ∩ {Y = y}) = P(X = x, Y = y) x, y ∈ R

Answer 41

ₓΣᵧΣpₓ,ᵧ (x, y) = 1

Answer 42

pₓ(x) = ᵧΣpₓ,ᵧ (x, y)

Answer 43

pᵧ(y) = ₓΣpₓ,ᵧ (x, y)

Answer 44

pᵧ|ₓ₌ₓ(y) = P (Y = y|X = x) | = pₓ,ᵧ(x,y)/pₓ(x) for y ∈ R

Answer 45

E [Y |X = x] = ᵧΣypᵧ|ₓ₌ₓ(y) | whenever the sum converges absolutely

Answer 46

P(X = x, Y = y) = P(X = x)P(Y = y) for all x, y ∈ R. In other words, X and Y are independent if and only if the events {X = x} and {Y = y} are independent for all choices of x and y. We can also write this as pₓ,ᵧ (x, y) = pₓ(x)pᵧ(y) for all x, y ∈ R

Answer 47

E[h(X, Y )] = ₓΣᵧΣ h(x, y)P(X = x, Y = y) = ₓΣᵧΣ h(x, y)pₓ,ᵧ (x, y) provided the sum converges absolutely.

Answer 48

E[aX + bY ] = aE[X] + bE[Y ] provided that both E [X] and E [Y ] exist. Prove it pg28

Answer 49

expectation is linear

Answer 50

E[a₁X₁ + · · · + aₙXₙ] = a₁E[X₁] + · · · + aₙE[Xₙ]

Answer 51

E[XY] = E[X]E[Y ] Proof pg28

Answer 52

cov (X, Y ) = E[(X − E [X])(Y − E [Y ])]

Answer 53

cov (X, X) = var (X)

Answer 54

pX₁,X₂,...,Xₙ (x₁, x₂, . . . , xₙ) = P(X₁ = x₁, X₂ = x₂, ..., Xₙ = xₙ) for x₁, x₂, ...,xₙ ∈ R

Answer 55

A family {Xᵢ : i ∈ I} of discrete random variables are independent if for all finite sets J ⊆ I and all collections {Aᵢ : i ∈ J} of subsets of R, P(ᵢ∈ⱼ∩{Xᵢ ∈ Aᵢ}) = ᵢ∈ⱼΠP(Xᵢ ∈ Aᵢ)

Answer 56

Independent and identically distributed (i.i.d)

Answer 57

ᵏΣⱼ₌₀ aⱼ uₙ₊ⱼ = f(n) with a₀ ≠ 0 and aₖ ≠ 0, where a₀...aₖ re constants independent of n A solution to such a difference equation is a sequence (uₙ)ₙ ≥ ₀ satisfying the sum for all n ≥ 0.

Answer 58

uₙ = vₙ +wₙ where (vₙ)ₙ ≥ ₀ is a particular solution to the equation and (wₙ)ₙ ≥ ₀ solves the homogeneous equation ᵏΣⱼ₌₀ aⱼ wₙ₊ⱼ = 0 proof pg31

Answer 59

Substitute wₙ = Aλⁿ in wₙ₊₁ + awₙ + bwₙ₋₁ = 0 then divide by Aλⁿ⁻¹ to get the quadratic: λ² + aλ + b = 0 (Aux Eqn) General Soln = wₙ = A₁λ₁ⁿ + A₂λ₂ⁿ or if λ₁ = λ₂ = λ then wₙ = (A + Bn)λⁿ

Answer 60

uₙ = { (q/p)ⁿ if p>q 1 if p ≤ q Proof pg 38

Answer 61

Gₓ(s) = E[sˣ] = ∞Σₖ₌₀ sᵏP(X=k)

Answer 62

pₓ(k) = pₖ = P(X=k)

Answer 63

Gₓ(s) = ₖΣpₖsᵏ = qs⁰ + ps¹ = q + ps | for all s ∈ R

Answer 64

Gₓ(s) = ⁿΣₖ₌₀ sᵏ ⁿCₖ pᵏ (1-p)ⁿ⁻ᵏ = ⁿΣₖ₌₀ ⁿCₖ (ps)ᵏ (1-p)ⁿ⁻ᵏ = (1 - p + ps)ⁿ by the binomial theorem. This is valid for all s ∈ R

Answer 65

Gₓ(s) = ∞Σₖ₌₀ sᵏ λᵏe^-λ/k! = e^-λ ∞Σₖ₌₀ (sλ)ᵏ/k! = e^λ(s-1) | for all s ∈ R

Answer 66

Gₓ(s) = ps/(1-(1-p)s) | provided that |s| < 1/1−p

Answer 67

Gₓ₊ᵧ(s) = Gₓ(s)Gᵧ(s)

Answer 68

Gₓ₊ᵧ(s) = E[sˣ⁺ʸ] = E[sˣsʸ] Since X and Y are independent, sˣ and sʸ are independent. So this equals E[sˣ]E[sʸ] = Gₓ(s)Gᵧ(s)

Answer 69

Y ∼ Bin(n, p)

Answer 70

Gᵧ(s) = E[sʸ] = E[s^(X₁ + ... + Xₙ)] = E[s^X₁] ... E[s^Xₙ] = (1 - p + ps)ⁿ As Y has the same p.g.f. as a Bin(n, p) random variable, we deduce that Y ∼ Bin(n, p).

Answer 71

ⁿΣᵢ₌₁ Xᵢ ∼ Po(ⁿΣᵢ₌₁ λᵢ) λᵢ = λ for all 1 ≤ i ≤ n: ⁿΣᵢ₌₁ Xᵢ ∼ Po(nλ) Proof pg41

Answer 72

``` G'ₓ(s) = d/ds E[sˣ] = d/ds ∞Σₖ₌₀ sᵏ P(X=k) = ∞Σₖ₌₀ d/ds sᵏ P(X=k) = ∞Σₖ₌₀ ksᵏ⁻¹P(X=k) = E[Xsˣ⁻¹] G'ₓ(1) = E[X] ```

Answer 73

G''ₓ(1) = E[X(X − 1)] = E[X²] − E[X],

Answer 74

var(X) = G''ₓ(1) + G'ₓ(1) - (G'ₓ(1))²

Answer 75

dᵏ/dsᵏ Gₓ(s) |ₛ₌₁ = E[X(X-1) ... (X - k + 1)]

Answer 76

The pgf of ᵢ₌₁Σᴺ Xᵢ is Gₙ(Gₓ(s)) Note that the sum ᵢ₌₁Σᴺ Xᵢ has a random number of terms. We interpret it as 0 if N = 0. Proof pg 44

Answer 77

ᵢ₌₁Σᴺ Xᵢ ∼ Po(λp)

Answer 78

Gₓ(s) = 1 - p + ps and Gₙ(s) = exp(λ(s − 1)) and so E[s^( ᵢ₌₁Σᴺ Xᵢ)] = Gₙ(Gₓ(s)) = exp(λ(1 - p + ps - 1)) = exp(λp(s-1)) Since this is the p.g.f. of Po(λp) and p.g.f.’s uniquely determine distributions, the result follows

Answer 79

Suppose we have a population (say of bacteria). Each individual in the population lives a unit time and, just before dying, gives birth to a random number of children in the next generation. This number of children has probability mass function p(i), i ≥ 0, called the offspring distribution

Answer 80

Xₙ₊₁ = C₁⁽ⁿ⁾ + C₂⁽ⁿ⁾ + ... + Cₓₙ⁽ⁿ⁾ We interpret this sum as 0 if Xₙ = 0 Note that C₁⁽ⁿ⁾, C₂⁽ⁿ⁾, .... are independent and identically distributed.

Answer 81

``` G(s) = ∞Σᵢ₌₀ p(i)sᶦ Gₙ(s) = E[sˣⁿ] (That's X subscript n) ```

Answer 82

Gₙ₊₁(s) = Gₙ(G(s)) = G(G(...G(s)...)) = G(Gₙ(s)) ^(n+1) times Proof pg 45

Answer 83

E[Xₙ] = µⁿ Proof pg 46

Answer 84

P(population dies out) = P(∞∪ₙ₌₀ {Xₙ = 0}) ≥ P (X₁ = 0) = p(0) > 0

Answer 85

Extinction Probability (non-examinable)

Answer 86

A random variable X defined on a probability space (Ω, F, P) is a function X : Ω → R such that {ω : X(ω) ≤ x} ∈ F for each x ∈ R.

Answer 87

is the function Fₓ : R → [0, 1] defined by Fₓ(x) = P (X ≤ x)

Answer 88

No, it's non-decreasing | Proof pg 51

Answer 89

P (a < X ≤ b) = Fₓ(b) − Fₓ(a) for a < b | Proof pg 51

Answer 90

x → −∞, Fₓ(x) → 0 | Proof pg 51/52

Answer 91

x → ∞, Fₓ(x) → 1 Proof pg x → ∞, Fₓ(x) → 151/52

Answer 92

Right Continuity

Answer 93

Fₓ(x) = P (X ≤ x) = −∞ ∫ˣ fₓ(u) du Bounds on the integral -∞ → x where fₓ : R → R is a function such that a) fₓ(u) ≥ 0 for all u ∈ R b) −∞ ∫ ∞ fₓ(u) du = 1

Answer 94

fₓ is called the probability density function (p.d.f.) of X or, sometimes, just its density.

Answer 95

dFₓ(x)/dx = fₓ(x) | at any point x such that fₓ(x) is continuous.

Answer 96

No!!!!! | Therefore it can exceed 1

Answer 97

P(X=x) = 0 for all x ∈ R | P(a ≤ X ≤ b) = ₐ∫ᵇ fₓ(x) dx

Answer 98

fₓ(x) = {1/b-a for a ≤ x ≤ b, | { 0 otherwise

Answer 99

X ∼ U[a, b]

Answer 100

fₓ(x) = λe^(-λx), x ≥ 0

Answer 101

α > 0 and λ ≥ 0 fₓ(x) = ((λ^α)/Γ(α)) x^(α-1)e^(-λx), x ≥ 0 Here, Γ(α) is the so-called gamma function, which is defined by Γ(α) = ∞∫₀ u^(α-1)e⁻ᵘ du for α > 0 For most values of α this integral does not have a closed form. However, for a strictly positive integer n, we have Γ(n) = (n − 1)!.

Answer 102

µ ∈ R and σ²> 0 fₓ(x) = 1/√2πσ² exp(-(x − µ)²/2σ² ), x ∈ R

Answer 103

X ∼ Gamma(α, λ)

Answer 104

X ∼ N(µ, σ²)

Answer 105

X ∼ N(µ, σ²)

Answer 106

P (x ≤ X ≤ x + δ) ≈ fₓ(x) δ

Answer 107

P (nδ ≤ X ≤ (n + 1)δ) ≈ fₓ(nδ)δ

Answer 108

E [X] = −∞ ∫ ∞ xfₓ(x) dx | whenever −∞ ∫ ∞ |x|fₓ(x) dx < ∞

Answer 109

E [h(X)] = −∞ ∫ ∞ h(x)fₓ(x) dx whenever −∞ ∫ ∞ |h(x)|fₓ(x) dx < ∞

Answer 110

E [aX + b] = aE [X] + b var (aX + b) = a²var (X) Proof pg 58

Answer 111

fᵧ(y) = fₓ(h⁻¹(y))d/dy h⁻¹(y) where h⁻¹ is the inverse function of h Proof pg60

Answer 112

Fₓ,ᵧ (x, y) = P (X ≤ x, Y ≤ y)

Answer 113

Fₓ,ᵧ(x, y) = 1

Answer 114

Fₓ,ᵧ(x, y) = 0

Answer 115

a) fₓ,ᵧ(u, v) ≥ 0 for all u, v ∈ R | b) −∞∫ʸ −∞∫ˣ fₓ,ᵧ(u, v) dudv = 1

Answer 116

their joint density function.

Answer 117

fₓ,ᵧ(x, y) = ∂²/∂x∂y Fₓ,ᵧ(x,y)

Answer 118

P (X ∈ A) = ₐ∫ fₓ(x) dx

Answer 119

P ((X, Y ) ∈ B) = ∫∫₍ₓ,ᵧ₎∈ᵦ fₓ,ᵧ(x, y)) dxdy

Answer 120

P (a < X ≤ b, c < Y ≤ d) = 𝒸∫ᵈ ₐ∫ᵇ fₓ,ᵧ(x, y)) dxdy for a < b and c < d Proof pg62

Answer 121

-∞∫∞ fₓ,ᵧ(x, y)) dy

Answer 122

-∞∫∞ fₓ,ᵧ(x, y)) dx Proof pg 63

Answer 123

Marginal distribution

Answer 124

fₓ,ᵧ(x, y) = fₓ(x) fᵧ(y) | for all x, y ∈ R

Answer 125

fₓ₁,ₓ₂,...,ₓₙ(x₁, x₂, . . . , xₙ) = fₓ₁(x₁)fₓ₂(x₂) ... fₓₙ(xₙ) for all x₁, x₂, . . . , xₙ∈ R

Answer 126

Fₓ,ᵧ (x, y) = Fₓ(x)Fᵧ(y) | for all x, y ∈ R.

Answer 127

E [h(X, Y )] = -∞∫∞ -∞∫∞ h(x, y) fₓ,ᵧ(x, y)) dxdy

Answer 128

cov (X, Y ) = E [(X − E [X])(Y − E [Y ])] = E [XY ] − E [X] E [Y ]

Answer 129

random sample of size n

Answer 130

_ | Xₙ = 1/n ᵢ₌₁Σⁿ Xᵢ

Answer 131

var (X + Y ) = var (X) + var (Y ) + 2cov (X, Y )

Answer 132

var(ᵢ₌₁Σⁿ Xᵢ) = ᵢ₌₁Σⁿ var(Xᵢ) + ᵢ≠ⱼΣcov(Xᵢ, Xⱼ) | = ᵢ₌₁Σⁿ var(Xᵢ) +2ᵢ

Answer 133

_ _ E[Xₙ] = µ and var(Xₙ) = 1/n σ² Proof pg 67

Answer 134

``` E[Xᵢ] = p var(Xᵢ) = p(1-p) for all 1 ≤ i ≤ n Hence, _ _ E[Xₙ] = p and var(Xₙ) = p(1-p)/n ```

Answer 135

X ∼ Ber(p) and E [X] = p

Answer 136

Suppose that X₁, X₂, . . . . are independent and identically distributed random variables with mean µ. Then for any fixed ε > 0 As n → ∞ P(|1/n ᵢ₌₁Σⁿ Xᵢ − µ| > ε)→0 Proof pg 68

Answer 137

P(|1/n ᵢ₌₁Σⁿ Xᵢ − µ| ≤ ε)→1

Answer 138

Suppose that Y is a non-negative random variable whose expectation exists. Then P(Y ≥ t) ≤ E[Y]/t for all t > 0. Proof pg68

Answer 139

Suppose that Z is a random variable with a finite variance. Then for any t > 0, P (|Z − E [Z] | ≥ t) ≤ var (Z)/t² Proof: Note that P (|Z − E [Z]| ≥ t) = P((Z − E [Z])² ≥ t²) and then apply Markov’s inequality to the non-negative random variable Y = (Z − E [Z])²

Probability Flashcards

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