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Example Problems on Multivariate Distributions

ST701 Homework 5 on Multivariate Distributions

Problems: 4.5, 4.10, 4.15, 4, 4.17, 4.24, 4.30, 4.40

4.5

a

Find P(X>Y) if X and Y are jointly distributed with pdf

f(x,y)=x+y, 0x1, 0y1. P(X>Y)=1y=01x=yf(x,y)dxdy=1y=01x=yx+ydxdy=1y=012x2+yx|1ydy=1y=012+y12yy3/2dy=12y+y2/425y5/2|10=720

b

Find P(X2<Y<X) if X and Y are jointly distributed with pdf

f(x,y)=2x, 0x1, 0y1. P(X2<Y<X)=1x=0xy=x22xdydx=1x=02xy|1x2dx=1x=02x22x3dx=23x324x4|10=16

4.10

The random pair (X,Y) has the distribution

X123211216112Y31601640130

a

Show that X and Y are dependent.

Notice that P(X=2,Y=3)=0, however P(X=2)P(Y=3)=26612=130. The product of the marginals is not the same probability of the joint, so they are not independent.

b

Given a probability table for random variables U and V that have the same marginals as X and Y but are independent.

X1232112212112412Y31122121124124112212112412312612312

4.15

Let XPoisson(θ), YPoisson(λ), independent. It was shown in Theorem 4.3.2 that the distribution of X+Y is Poisson(θ+λ). Show that the distribution of X|X+Y is binomial with success of probability θ/(θ+λ). What is the distribution of Y|X+Y?

\begin{align} f_{X|X+Y} & = \frac{ f(X, X+Y) }{ f(X+Y) } \\ & = \frac{ \frac{ e^{-\theta} \theta^x }{ x! } \frac{e^{\-\lambda} \lambda^y }{ y! } }{ \frac{ e^{-(\theta+\lambda)} (\theta + \lambda)^{x+y} }{ (x+y)!} }\\ & = \frac{ (x+y)! }{ x! y! } \frac{ e^{-\theta} \theta^x e^{\-\lambda} \lambda^y }{ e^{-(\theta+\lambda)} (\theta + \lambda)^{x+y} } \\ & = \frac{ (x+y)! }{ x! y! } \frac{ \theta^x \lambda^y }{ (\theta + \lambda)^{x+y} } \\ & = {x+y \choose x} \Big( \frac{ \theta }{ \theta + \lambda} \Big)^x \Big( \frac{ \lambda }{ \theta + \lambda } \Big)^y \end{align}

Taking p = \Big( \frac{ \theta }{ \theta + \lambda} \Big)^x and 1 - p = \Big( \frac{ \lambda }{ \theta + \lambda } \Big)^y, we get X|X+Y \sim Binomial( p = \frac{ \theta }{ \theta + \lambda }). Similarly, Y | X = Y \sim Binomial( p = \frac{ \lambda }{ \theta + \lambda }).

4

If X has a binomial distribution with parameters n and p, Y has a binomial distribution with parameters m and p, and X, Y are independent. Show that X + Y has a binomial distribution with parameters n + m and p.

Because the random variables are independent, we can use the product of the MGF to find the MGF of X+Y.

\begin{align} M_Z(t) & = M_X(t) M_Y(t) \\ & = \Big[ p e^t + (1-p) \Big]^n \Big[ p e^t + (1-p) \Big]^m \\ & = \Big[ p e^t + (1-p) \Big]^{n+m} \\ \end{align}

Thus, Z \sim Bin(n+ m, p).

4.17

Let X be an exponential(1) random variable and define Y to be the integer part of X+1, that is

Y = i + 1\ \text{if and only if} \ i \leq X < i + 1, \ i = 0, 1, 2, \dots

a

Find the distribution of Y. What well-known distribution does Y have?

\begin{align} Y & = \int_{i}^{i+1} e^{-x} dx \\ & = -e^{-(1+i - i)} \\ & = -e^{-(i+1) + e^{-i}} \\ & = e^{-i} - e^{-i} e^{1} \\ & = e^{-i}(1-e) \\ & = e^{-1 (i)} (1-e^{-1}) \end{align}

Notice that 1-e^{-1} = p. Thus, Y \sim Geometric(1-e^{-1}).

b

Find the conditional distribution of X - 4 given Y \geq 5.

\begin{align} P(X-4 \leq x | Y \leq 5) & = P(X - 4 \leq x | X \geq 4) \\ & = \frac{ P(X - 4 \leq x) \land X \geq 4 }{ P(X \geq 4) } \\ & = \frac{ P(X \leq x+4 \land X \geq 4) }{ P(X \geq 4) } \\ & = \frac{ P(4 \leq X \leq x-4) }{ P(X \geq 4) } \\ & = \frac{ \int_{4}^{x+4} e^{-z} dz }{ \int_{4}^{\inf} e^{-x} dx } \\ & = \frac{ -(e^{-(x+4)} - e^{-4}) }{ - (e^{-\inf} - e^{-4}) } \\ & = \frac{ e^{-4} - e^{-(x+4)} }{ e^{-4} } \\ & = 1 - e^{-x} \end{align}

This follows an Exp(1) distribution.

4.24

Let X and Y be independent random variables with X \sim gamma(r, 1) and Y \sim gamma(s,1). Show that Z_1 = X+Y and Z_2 = X/(X+Y) are independent, and find the distribution of each (Z_1 is gamma and Z_2 is beta).

Notice that X = Z_1 Z_2 and Y = Z_1 - Z_1 Z_2 = Z_1(1-Z_2). We can find the determinate of the Jacobian.

det \begin{bmatrix} Z_2 & Z_1 \\ 1-Z_2 & -Z_1 \end{bmatrix} = -Z_2 Z_1 (Z_1 - Z_1 Z_2) = -Z_1 \begin{align} f_{Z_1, Z_2} & = f(X, Y) \cdot | J | \\ & = f_X(Z_1 Z_2) f_Y(Z_1(1-Z_2)) \cdot Z_1 \\ & = \frac{ 1 }{ \Gamma(R) 1^r } (z_1 z_2)^{r-1} e^{-(z_1 z_2) / 1} \cdot \frac{ 1 }{ \Gamma(s) 1^s } (z_1-z_1 z_2)^{s-1} e^{-(z_1-z_1 z_2)/2}z_1 \\ & = \frac{ 1 }{ \Gamma(r) } z_1^{r-1} z_2^{r-1} \frac{ 1 }{ \Gamma(s) } z_1^{s-1} (1-z_2)^{s-1} e^{-z_1} z_1 \\ & = \frac{ 1 }{ \Gamma(r+s) } z_1^{r+s-1} e^{-z_1} \cdot \frac{ \Gamma(r+s)}{\Gamma(r)\Gamma(s)} z_2^{r-1} (1-z_2)^{s-1} \end{align}

Notice that we can separate the z_1 and z_2 terms, so they are independent. Also, Z_1 \sim Gamma(r+s,1) and Z_2 \sim Beta(r,s)

4.30

Suppose the distribution of Y, conditional on X=x is n(x,x^2) and that the marginal distribution of X is uniform(0,1).

a

Find E(Y), Var(Y), and Cov(X,Y).

\begin{align} E(Y) & = E(E(Y|X)) \\ & = E(E(N(X, X^2))\\ & = E(X) \\ & = \frac{ 1 }{ 2 } \end{align} \begin{align} Var(Y) & = Var(E(Y|X)) + E(Var(Y|X)) \\ & = E(X^2) + Var(X) \\ & = Var(X) + E(X)^2 + Var(X) \\ & = \frac{ 1 }{ 12 } + (\frac{ 1 }{ 2 })^2 + \frac{ 1 }{ 12 } \\ & = \frac{ 5 }{ 12 } \end{align} \begin{align} Cov(X, Y) & = E(XY) - E(X)E(Y) \\ & = E(E(XY|X)) - \frac{ 1 }{ 2 } \frac{ 1 }{ 2 }\\ & = E(X E(Y|X) ) - \frac{ 1 }{ 4} \\ & = E(X \cdot X) - \frac{ 1 }{ 4} \\ & = Var(X) + E(X)^2 - \frac{ 1 }{ 4 } \\ & = \frac{ 1 }{ 12 } + \frac{ 1 }{ 4 } - \frac{ 1 }{ 4 } \\ & = \frac{ 1 }{ 12 } \end{align}

b

Prove that Y/X and X are independent.

Take U = Y/X and V = X. So, X = V and Y = UV. Then,

det \begin{bmatrix} 0 & 1 \\ V & U \end{bmatrix} = -V \begin{align} f_{U,V} & = f(x, y) \cdot | J | \\ & = f_{y|x} \cdot f_X \cdot | -v | \\ & = \frac{ 1 }{ \sqrt{2 \pi x^2 } } e^{-\frac{ (y-x)^2 }{ 2x^2 }} \cdot \frac{ 1 }{ 1-0 } v \\ & = \frac{ 1 }{ \sqrt{2 \pi v^2} } e^{-\frac{ (uv-v)^2 }{ 2v^2 }} v \\ & = \frac{ v }{ \sqrt{2 \pi v^2} } e^{\frac{ -u^2 v^2 + 2 uv^2 - v^2 }{ 2 v^2 }} \\ & = \frac{ v }{ \sqrt{2 \pi v^2} } \cdot e^{\frac{ -u^2+2u + -1 }{ 2 }} \end{align}

Notice that the left expression is f_X and the right expression is f_{Y/X}. Since we can separate the terms in the joint distribution, they are independent.

4.40

A generalization of the beta distribution is the Dirichlet distribution. In its bivariate version (X,Y) has pdf

f(x,y) = C x^{a-1} y^{b-1}(1-x-y)^{c-1}, \ 0 < x < 1, \ 0 < y < 1, \ 0 < y < 1 - x < 1,

where a>0, b>0, and c > 0 are constants.

a

Show that C = \frac{ \Gamma(a + b + c) }{ \Gamma(a) \Gamma(b) \Gamma(c) }.

Take z = \frac{ y }{ 1-x } and dz = \frac{ 1 }{ 1-x } dy.

\begin{align} 1 & = \int_{x=0}^{1} \int_{y=0}^{1-x} C x^{a-1} y^{b-1} (1-x-y)^{c-1} dy dx \\ 1/C & = \int_{x=0}^{1} \int_{y=0}^{1-x} x^{a-1} y^{b-1} (1-x-y)^{c-1} dy dx \\ & = \int_{x=0}^{1} \int_{z=0}^{1} x^{a-1} (z(1-x)^{b-1} (1-x-(z (1-x)))^{c-1} (1-x)dz dx \\ & = \int_{x=0}^{1} x^{a-1} (x-1)^{c+b-2} (x-1) dx \int_{z=0}^{1} (z)^{b-1} (1-z)^{c-1} dz \\ & = \frac{ \Gamma(a) \Gamma(c+b) }{ \Gamma(a+b+c) } \int_{x=0}^{1} \frac{ \Gamma(a+c) }{ \Gamma(a) \Gamma(c) } x^{a-1} (x-1)^{c+b-2} (x-1) dx \cdot \frac{ \Gamma(b) \Gamma(c) }{ \Gamma(b+c) }\int_{z=0}^{1} \frac{ \Gamma(b+c) }{ \Gamma(b) \Gamma(c)} (z)^{b-1} (1-z)^{c-1} dz \\ \\ C & = \frac{ \Gamma(a + b + c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } \end{align}

b

Show that, marginally, both X and Y are beta.

We will use z = \frac{ x }{ 1-y } and dz = \frac{ 1 }{1-y } dx.

\begin{align} f_Y & = \int_{x=0}^{1-y} \frac{ \Gamma(a+b+c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } x^{a-1} y^{b-1} (1-x-y)^{c-1} dx \\ & = \frac{ \Gamma(a+b+c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } y^{b-1} \int_{z=0}^{1} (z(1-y))^{a-1} (1-z(1-y)-y)^{c-1} (1-y) dz \\ & = \frac{ \Gamma(a+b+c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } y^{b-1} (1-y)^{a-1} (1-y)^{c-1} (1-y) \int_{z=0}^{1} (z))^{a-1} (1-z)^{c-1} dz \\ & = \frac{ \Gamma(a+b+c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } y^{b-1} (1-y)^{a+c-1} \cdot \frac{ \Gamma(a) \Gamma(c) }{ \Gamma(a+c) } \int_{z=0}^{1} \frac{ \Gamma(a+c) }{ \Gamma(a)\Gamma(c) } (z)^{a-1} (1-z)^{c-1} dz \\ & = \frac{ \Gamma(a+b+c) }{ \Gamma(a+c) \Gamma(b) } y^{b-1} (1-y)^{a+c-1} \end{align}

Thus Y \sim Beta(b, a+c). The same process gives X \sim Beta(a, b+c).

c

Find the conditional distribution of Y| X = x, and show that Y/(1-x) is beta(b,c).

\begin{align} f_{Y|X } & = \frac{ \frac{ \Gamma(a+b+c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } x^{a-1} y^{b-1} (1-x-y)^{c-1} }{ \frac{ \Gamma(a+b+c) }{ \Gamma(a+c) \Gamma(b) } x^{a-1} (1-x)^{b+c-1} } \\ & = \frac{ \Gamma(b+c) }{ \Gamma(b) \Gamma(c) } y^{b-1} \frac{ (1-x-y)^{c-1} }{ (1-x)^{b+c-1} } \end{align}

Take U = Y/(1-X) and V = X. Then, X=V and Y=U(1-V).

det \begin{bmatrix} 1-V & -U \\ 0 & 1 \end{bmatrix} = -(1-V) \begin{align} f_{U,V} & = f(X,Y) |J| \\ & = f(V, U(1-V)) (1-V) \\ & = \frac{ \Gamma(a+b+c) }{ \Gamma(a)\Gamma(b)\Gamma(c) } v^{a-1} (u(1-v))^{b-1} (1-v-u(1-v))^{c-1}(1-v) \\ & = \frac{ \Gamma(a+b+c) }{ \Gamma(a)\Gamma(b)\Gamma(c) } v^{a-1} (1-v)^{b-1} (1-v)^{c-1} u^{b-1} (1-u)^{c-1} (1-v) \\ \\ f_U & = \frac{ \Gamma(a+b+c) }{ \Gamma(a)\Gamma(b)\Gamma(c) } u^{b-1} (1-u)^{c-1} \cdot \int_{v=0}^{1} v^{a-1} (1-v)^{b+c-1} dv \\ & = \frac{ \Gamma(a+b+c) }{ \Gamma(a)\Gamma(b)\Gamma(c) } u^{b-1} (1-u)^{c-1} \cdot \frac{ \Gamma(b+c) \Gamma(a) }{ \Gamma(a+b+c) } \int_{v=0}^{1} \frac{ \Gamma(a+b+c) }{ \Gamma(a) \Gamma(b+c) }v^{a-1} (1-v)^{b+c-1} dv \\ & = \frac{ \Gamma(b+c) }{ \Gamma(b) \Gamma(c) } u^{b-1} (1-u)^{c-1} \end{align}

Thus, Y / (1-X) \sim Beta(b, c)

d

Show that E(XY) = \frac{ ab }{ (a+b+c+1)(a+b+c)}, and find their covariance.

\begin{align} E(XY) & = \int_{x=0}^{1} \int_{y=0}^{1-x} x y \cdot \frac{ \Gamma(a+b+c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } x^{a-1} y^{b-1} (1-x-y)^{c-1} dy dx \\ & = \frac{ \Gamma(a+b+c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } \frac{\Gamma(a+1) \Gamma(b+1) \Gamma(c) }{ \Gamma(a + 1 + b + 1 + c) } \int_{x=0}^{1} \int_{y=0}^{1-x} \frac{ \Gamma(a+1 + b + 1+c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } x^{a+1-1} y^{b+1-1} (1-x-y)^{c-1} dy dx \\ & = \frac{ a \Gamma(a) b \Gamma(b) \Gamma(c) }{ \Gamma(a) \Gamma(b) \Gamma(c) } \cdot \frac{ \Gamma(a+b+c) }{ (a+b+c+1)(a+b+c) \Gamma(a+b+c) } \\ & = \frac{ ab }{ (a+b+c+1)(a+b+c) } \\ \\ Cov(X, Y) & = E(XY) - E(X) E(Y) \\ & = \frac{ ab }{ (a+b+c+1)(a+b+c) } - \frac{ a }{ a+ b + c } \frac{ b }{ a + b + c } \\ & = ab (\frac{ 1 }{(a+b+c+1)(a+b+c) } - \frac{ 1 }{ (a+ b + c )^2 }) \end{align}