Solutions

2.5

$X \sim U[0,2\pi]$ and $Y = \sin^{2}(X)$. Write

$$\sin^{2}(x) = \begin{cases}\frac{1}{2} + \frac{1}{2}\sin 2\left(x... ...\sin 2\left(x-\frac{7}{4}\pi\right) & \frac{3}{2}\pi < x \le 2\pi \end{cases}$$

So

$$g_{1}^{-1}(y) = \frac{1}{2} (2y-1)+\frac{1}{4}\pi$$

$$g_{2}^{-1}(y) = \frac{1}{2} (-2y+1)+\frac{3}{4}\pi$$

$$g_{3}^{-1}(y) = \frac{1}{2} (2y-1)+\frac{5}{4}\pi$$

$$g_{4}^{-1}(y) = \frac{1}{2} (-2y+1)+\frac{7}{4}\pi$$

and

$$\left\vert\frac{d}{dy}g_{i}^{-1}(y)\right\vert = \frac{1}{\sqrt{1-(2y-1)^{2}}}$$

So for $0 < y < 1$

$$f_{Y}(y) = 4 \frac{1}{2\pi} \frac{1}{\sqrt{1-(2y-1)^{2}}}$$

$$= \frac{2}{\pi} \frac{1}{2 \sqrt{y} \sqrt{1-y}}$$

$$= \frac{1}{\pi\sqrt{y-y^{2}}}$$

For the alternative approach,

$$F_{Y}(y) = 2 P(X \le x_{1}) + 2 P(x_{2} \le X \le \pi)$$

Since $\sin x_{i} = y$,

$$\left\vert\frac{d x_{i}}{dy}\right\vert = \frac{1}{2 \sqrt{y-y^{2}}}$$

and thus

$$F_{Y}'(y) = \frac{4}{2\pi} \frac{1}{2\sqrt{y-y^{2}}} = \frac{1}{\pi\sqrt{y-y^{2}}}$$

2.8

$F^{-1} = \inf\{x: F(x) \ge y\}$

a.

$$F(x) = \begin{cases}0 & x < 0 1-e^{-x} & x \ge 0 \end{cases} F^{-1}(y) = \begin{cases}-\infty & y=0 -\log(1-y) & y > 0 \end{cases}$$

b.

$$F(x) = \begin{cases}\frac{1}{2}e^{x} & x < 0 \frac{1}{2} & 0 \le x < 1 1-\frac{1}{2}e^{1-x} & x \ge 1 \end{cases} F^{-1}(y) = \begin{cases}\log 2y & 0 \le y \le 1/2 1-\log(2(1-y)) & 1/2 < y \le 1 \end{cases}$$

c.

$$F(x) = \begin{cases}\frac{1}{4} e^{x} & x < 0 1-\frac{1}{4} e^{-x} & x \ge 0 \end{cases} F^{-1}(y) = \begin{cases}\log (4y) & 0 \le y < 1/4 0 & 1/4 \le y < 3/4 -\log(4(1-y)) & 3/4 \le y \le 1 \end{cases}$$

2.9

L^ATEX2Html is giving me grief about this one-it is available in the pdf version.

2.11

a.

$$E[X^{2}] = \int_{-\infty}^{\infty}x^{2}\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}dx$$

$$= \int_{-\infty}^{\infty}x\frac{x}{\sqrt{2\pi}}e^{-x^{2}/2}dx$$

$$=\left.-x\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}\right\vert _{-\infty}^{\infty} + \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}dx$$

$$= 1$$

or

$$E[X^{2}] = E[Y] = \int_{0}^{\infty}y\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{y}}e^{-y/2}dy$$

$$= \frac{1}{2\pi}\int_{0}^{\infty}\sqrt{y}e^{-y/2}dy$$

$$= \frac{2^{3/2}}{\sqrt{2\pi}}\int_{0}^{\infty}z^{3/2-1}e^{-z}dz$$

$$= \frac{2^{3/2}}{\sqrt{2\pi}}\Gamma(3/2) = \frac{2}{\sqrt{\pi}}\Gamma(3/2)$$

$$= \frac{2}{\sqrt{\pi}}\Gamma(1/2)\frac{1}{2} = \frac{\Gamma(1/2)}{\sqrt{\pi}} = 1$$

b.

$$f_{Y}(y) = \begin{cases}2 e^{-y^{2}/2}\frac{1}{2\sqrt{y}} & y > 0 0 & y < 0 \end{cases}$$

$$E[Y] = E[\vert X\vert] = 2\int_{0}^{\infty}ye^{y^{2}/2}\frac{1}{\sqrt{2\pi}}dy$$

$$= 2 \left[\left.-\frac{1}{\sqrt{2\pi}}e^{-y^{2}/2}\right\vert _{0}^{\infty} \right] = 2\frac{1}{\sqrt{2\pi}} = \sqrt{\frac{2}{\pi}}$$

$$E[Y^{2}] = E[X^{2}] = 1$$

$$(Y) = 1-\frac{2}{\pi}$$

2.14

(a)

$$E[X] = \int_{0}^{\infty}(1-F(x))dx$$

$$= \int_{0}^{\infty}1(1-F(x))dx$$

$$= \left.x(1-F(x))\right\vert _{0}^{\infty}+\int_{0}^{\infty}xf(x)dx$$

[ $x(1-F(x)) \le \int_{x}^{\infty}uf(u)du\rightarrow 0$ if $E[X]<\infty$]

(b)

Write the expectation as

$$E[X] = \sum_{i=0}^\infty i f_X(i) = \sum_{i=1}^\infty i f_X(i) = \sum_{i=1}^\infty \sum_{j=1}^i f_X(i)$$

replacing $i f_X(i)$ by a sum that adds up $f_X(i)$ $i$ times. Change the order of summation:

$$\sum_{i=1}^\infty \sum_{j=1}^i f_X(i) = \sum_{i=1}^\infty \sum_{j=i}^\infty f_X(i) = \sum_{j=1}^\infty P(X \ge j)$$

$$= \sum_{j=1}^\infty (1 - F_X(j-1)) = \sum_{k=0}^\infty (1 - F_X(k))$$

Since $F_X$ is a right-continuous step function with jumps at the integers,

$$1 - F_X(k) = \int_k^{k+1} 1 - F_X(x) dx$$

for $k = 0, 1, 2, \dots,$ so

$$E[X] = \int_0^\infty 1 - F_X(x) dx$$

in this case as well

The result holds for all non-negative random variables. A general derivation:

$$E[X] = E\left[\int_{0}^{X}1du\right]$$

$$= E\left[\int_{0}^{\infty}h(X,u)du\right]$$

where

$$h(a,b) = \begin{cases}1 & \text{if $a,b \ge 0$ and $a > b$} 0 & \text{otherwise} \end{cases}$$

Reverse order of integration:

$$E[X] = \int_{0}^{\infty}E[h(X,u)]du$$

For fixed $u$, $h(X,u) = 1$ if and only if $X > u$, so

$$E[h(X,u)] = P(X > u) = 1 - F(u)$$

So

$$E[X] = \int_{0}^{\infty} 1-F(u) du$$

for any $F$ where $X \ge 0$.

2.16

$P(T>t) = ae^{-\lambda t}+(1-a)e^{-\mu t}$ for $t>0$.

$$E[T] = \int_{0}^{\infty}P(T>t)dt = a\int_{0}^{\infty}e^{-\lambda t}dt + (1-a)\int_{0}^{\infty}e^{-\mu t}dt$$

$$= \frac{a}{\lambda}\int_{0}^{\infty} \lambda e^{-\lambda t} dt + \frac{1-a}{\mu} \int_{0}^{\infty} \mu e^{-\mu t} dt$$

$$= \frac{a}{\lambda} + \frac{1-a}{\mu}$$

2.24

a.

$f(x) = a x^{a-1}$, $0 < x < 1$, $a > 0$.

$$E[X] = \int_{0}^{1}a x^{a}dx = \frac{a}{a+1}$$

$$E[X^{2}] = \int_{0}^{1}a x^{a+1}dx = \frac{a}{a+2}$$

$$(X) = \frac{a}{a+2}-\left(\frac{a}{a+1}\right)^{2}$$

b.

$f(x) = \frac{1}{n}$, $x = 1, 2, \ldots, n$.

$$E[X] = \sum_{i=1}^{n}\frac{i}{n} = \frac{1}{n} \sum_{i=1}^{n}i = \frac{1}{n}\frac{n(n+1)}{2} =\frac{n+1}{2}$$

$$E[X^{2}] = \sum_{i=1}^{n}\frac{i^{2}}{n} =\frac{1}{n}\sum_{i=1}^{n}i^{2} =\frac{1}{n}\frac{n(n+1)(2n+1)}{6} =\frac{(n+1)(2n+1)}{6}$$

$$(X) = \frac{(n+1)(2n+1)}{6}-\left(\frac{n+1}{2}\right)^{2} = (n+1) \left(\frac{2n+1}{6}-\frac{n+1}{4}\right)$$

$$= (n+1) \frac{4n+2-3n-3}{12} =(n+1)\frac{n-1}{12} = \frac{n^{2}-1}{12}$$

c.

$f(x) = \frac{3}{2}(x-1)^{2}$, $0 < x < 2$.

$$E[X] = \int_{0}^{2}x\frac{3}{2}(x-1)^{2}dx = 1$$

$$(X) = E[(X-1)^{2}]$$

$$= \int_{0}^{2}\frac{3}{2}(x-1)^{4}dx$$

$$= \frac{3}{2}\left.\frac{(x-1)^{5}}{5}\right\vert _{0}^{2} = \frac{3}{2}\times\frac{2}{5} = \frac{3}{5}$$