Solutions

4.27

Approach from class: Let $Z_{1}, Z_{2}$ be independent standard normals and let

$$X = \mu + \sigma Z_{1}$$

$$Y = \gamma + \sigma Z_{2}$$

Then

$$U = X + Y = \mu + \gamma + \sigma Z_{1} + \sigma Z_{2}$$

$$V = X - Y = \mu - \gamma + \sigma Z_{1} - \sigma Z_{2}$$

So

$$C = B B^{T} = \begin{bmatrix}2 \sigma^{2} & 0 0 & 2 \sigma^{2} \end{bmatrix}$$

and

$$f_{U,V}(u,v) = \frac{1}{2\pi 2\sigma^{2}} \exp\left\{-\frac{(u-(\mu+\gamma))^{2}}{4\sigma^{2}} -\frac{(v-(\mu-\gamma))^{2}}{4\sigma^{2}}\right\}$$

$$= f_{U}(u) f_{V}(v)$$

where $U \sim N(\mu+\gamma,2\sigma^{2})$ and $V \sim N(\mu-\gamma,2\sigma^{2})$; $U, V$ are independent.

4.30

a.

The mean of $Y$ is

$$E[Y] = E[E[Y\vert X]] = E[X] = 1/2$$

The variance is

$$(Y) = E[ (Y\vert X)] + (E[Y\vert X]) = E[X^2] + (X)$$

$$= 1/3 + 1/12 = 5/12$$

The covariance is

$$(X,Y) = E[(Y-\mu_Y)(X-\mu_X)] = E[E[Y-\mu_Y\vert X](X-\mu_X)]$$

$$= E[(X - \mu_X)^2] = (X) = 1/12$$

b.

The conditional distribution of $Z = Y/X$, given $X = x$ is $N(1,1)$. Since this conditional distribution does not depend on $x$, $Z$ and $X$ are independent.

5.2

a.

Condition on $X_{1}=x$:

$$P(Y > y \vert X_{1} = x) = \begin{cases}1 & \text{if $y \le 0$} F(x)^{y} & \text{if $y \ge 1$} \end{cases}$$

i.e. $Y\vert X_{1}=x$ is geometric with $p = 1-F(x)$. So for $y \ge 1$

$$P(Y > y) = E[F(X_{1})^{y}] = \int_{0}^{1}u^{y}du$$

$$= \frac{1}{y+1}$$

since $F(X_{1})$ is uniform on $[0,1]$ by the probability integral transform. So for $y = 1, 2, \ldots,$

$$P(Y=y) = \frac{1}{y}-\frac{1}{y+1} = \frac{1}{y(y+1)}$$

Alternative argument: For $y = 1, 2, \ldots,$

$$P(Y > y) = P(X_{1} > \max\{X_{2}, \ldots, X_{y+1}\})$$

$$= P( ({X}_{1},\ldots,{X}_{y+1})=1)$$

$$= \frac{1}{y+1}$$

by symmetry.

b.

Using $\lfloor y \rfloor$ to denote the largest integer less than or equal to $y$ we have $P(Y > y) = 1 - F_Y(y) = 1/(\lfloor y \rfloor + 1)$ for all $y \ge 0$. So

$$E[Y] = \int_0^\infty 1 - F_Y(y) dy = \int_0^\infty \frac{1}{\lfloor y \rfloor + 1} dy$$

$$\ge \int_0^\infty \frac{1}{y + 1} dy = \infty$$

5.4

a.

Suppose $X_{i_{1}},\ldots,X_{i_{k}}\vert P$ are independent Bernoulli($P$). Then

$$P(X_{i_{1}}=x_{1},\ldots,X_{i_{k}}=x_{k}\vert P=p) = p^{\sum_{i=1}^{k}x_{i}}(1-p)^{\sum_{i=1}^{k}(1-x_{i})}$$

$$= p^{\sum_{i=1}^{k}x_{i}}(1-p)^{k-\sum_{i=1}^{k}x_{i}}$$

$$= p^{t}(1-p)^{k-t}$$

So

$$P(X_{i_{1}}=x_{1},\ldots,X_{i_{k}}=x_{k}) = \int_{0}^{1}p^{t}(1-p)^{k-t}dp$$

$$= \frac{\Gamma(t+1)\Gamma(k-t+1)}{\Gamma(k+2)}$$

$$= \frac{t!(k-t)!}{(k+1)!}$$

b.

$P(X_{1}=x_{1})=\frac{x_{1}!(1-x_{1})!}{2!} = \frac{1}{2}$. So

$$P(X_{1}=x_{1}) \times\cdots\times P(X_{k}=x_{k}) = \frac{1}{2^{k}}$$

For $k = 2$,

	(0,0)	(0,1)	(1,0)	(1,1)
$t!(2-t)!/3!$	1/3	1/6	1/6	1/3
independent	1/4	1/4	1/4	1/4