Solutions

3.23

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

3.24

a.

$Y=X^{1/\gamma}$, $X \sim$ Exponential($\beta$).

$$f_{Y}(y) = \begin{cases}\frac{\gamma}{\beta}y^{\gamma-1}e^{-y^{\gamma}/\beta} & y > 0 0 & y \le 0 \end{cases}$$

$$E[Y^{k}] = \int_{0}^{\infty}x^{k/\gamma}\frac{1}{\beta}e^{-x/\beta}dx = \Gamma\left(\frac{k}{\gamma}+1\right)\beta^{k/\gamma}$$

$$E[Y] = \Gamma(1/\gamma+1)\beta^{1/\gamma}$$

$$(Y) = \Gamma(2/\gamma+1)\beta^{2/\gamma} -\Gamma(1/\gamma+1)^{2}\beta^{2/\gamma}$$

b.

$Y=(2X)^{1/2}$, $X \sim$   Gamma$(1,1) =$   Exponential$(1)$.

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

$$E[Y] = \int_{0}^{\infty} \sqrt{2} \sqrt{x} e^{-x} dx = \sqrt{2}\int_{0}^{\infty}x^{3/2-1}e^{-x}dx = \sqrt{2}\Gamma(3/2)$$

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

$$E[Y^{2}] = 2 \int_{0}^{\infty}x e^{-x} dx = 2 \Gamma(2) = 2$$

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

c.

$X \sim$   Gamma$(a,b)$, $Y = 1/X$.

$$f_Y(y) = \frac{e^{-1/(\beta y)}}{\Gamma(\alpha) \beta^{\alpha} y^{\alpha+1}}$$

for $y > 0$. Moments:

$$E[Y^k] = E[X^{-k}] = \int_0^\infty \frac{x^{\alpha-k-1} e^{-\beta x}}{\Gamma(\alpha) \beta^\alpha} = \frac{\Gamma(\alpha-k)}{\Gamma(\alpha)\beta^k}$$

for $\alpha > k$ and $E[Y^{-k}] = \infty$ for $\alpha \le k$. So for $\alpha > 2$

$$E[Y] = \frac{\Gamma(\alpha-1)}{\Gamma(\alpha)\beta} = \frac{1}{(\alpha-1)\beta}$$

$$E[Y^2] = \frac{\Gamma(\alpha-2)}{\Gamma(\alpha)\beta^2} = \frac{1}{(\alpha-1)(\alpha-2)\beta^2}$$

$$(Y) = \frac{1}{(\alpha-1)\beta^2} \left(\frac{1}{\alpha-2}-\frac{1}{\alpha-1}\right) = \frac{1}{(\alpha-1)^2(\alpha-2)\beta^2}$$

d.

$X \sim$   Gamma$(3/2,1)$, $Y = X^{1/2}$.

$$f_{Y}(y) = \begin{cases}\frac{2}{\Gamma(3/2)}y^{2}e^{-y^{2}} & y > 0 0 & y \le 0 \end{cases}$$

$$E[Y^{k}] = E[X^{k/2}] = \int_{0}^{\infty}\frac{1}{\Gamma(3/2)}x^{k/2+3/2-1}e^{-x}dx$$

$$= \Gamma\left(\frac{k+3}{2}\right)/\Gamma(3/2)$$

$$E[Y] = \Gamma(2)/\Gamma(3/2) = 1/(\sqrt{\pi}/2) = \frac{2}{\sqrt{\pi}}$$

$$E[Y^{2}] = \Gamma(5/2)/\Gamma(3/2) = \frac{3}{2}$$

$$(X) = \frac{3}{2}-\frac{4}{\pi}$$

e.

$X \sim$   Exponential$(1)$, $Y = \alpha - \gamma \log X$

$$f_{Y}(y) = \frac{1}{\gamma}\exp\{e^{-(y-\alpha)/\gamma}-(y-\alpha)/\gamma\}$$

$$E[Y] = \alpha-\gamma E[\log X]$$

$$M_{\log X}(t) = E[e^{t \log X}] = E[X^{t}] = \int_{0}^{\infty}x^{(t+1)-1}e^{-x}dx = \begin{cases}\Gamma(t+1) & t > -1 \infty & t \le -1 \end{cases}$$

$$E[\log X] = \Gamma'(1) = -( ) = -0.577216$$

$$E[(\log X)^{2}] = \Gamma''(1) = ( )^{2} + \pi^{2}/6 = 1.97811$$

$$(\log X) = \Gamma''(1)-\Gamma'(1)^{2} = \pi^{2}/6$$

Once the moment generating function has been obtained, the rest can be done in Mathematica: The derivative of the Gamma function is obtained as

In[23]:= D[Gamma[x],x]

Out[23]= Gamma[x] PolyGamma[0, x]

The value at $x=1$ is obtained using the ``slash-dot'' operator:

In[24]:= % /. x->1

Out[24]= -EulerGamma

The variable % refers to the last output expression. The numerical value is obtained to 10 digits by

In[25]:= N[%,10]

Out[25]= -0.5772156649

The second derivative of the Gamma function is

In[26]:= D[Out[23],x]

                                 2
Out[26]= Gamma[x] PolyGamma[0, x]  + Gamma[x] PolyGamma[1, x]

Substituting $x=1$ produces

In[27]:= % /. x->1

                         2
                   2   Pi
Out[27]= EulerGamma  + ---
                        6

In[28]:= N[%,10]

Out[28]= 1.978111991

3.25

$$h_{T}(t) = \lim_{\delta \downarrow 0}\frac{P(1 \le T < t+\delta\vert T>t)}{\delta}$$

$$= \lim_{\delta \downarrow 0}\frac{1}{\delta} \frac{F(t+\delta)-F(t)}{1-F(t)}$$

$$= \frac{\frac{1}{\delta}(F(t+\delta)-F(t))}{1-F(t)}$$

$$= \frac{F'(t)}{1-F(t)} = \frac{f(t)}{1-F(t)}$$

and

$$- \frac{d}{dt}\log(1-F(t))=\frac{f(t)}{1-F(t)}$$

The quantity

$$H_{T}(t) = -\log(1-F_{T}(t)) = \int_{0}^{t}h_{T}(u)du$$

is called the cumulative hazard function, and

$$F_{T}(t) = 1-\exp\{-H_{T}(t)\}$$

Constant hazard: $h_{T}(t) \equiv c$.

$$F_{T}(t) = 1-e^{-ct}$$

an exponential distribution.

Power hazard: Weibull.

3.26

a.

$$f_{X}(t) = \frac{1}{\beta} e^{-t/\beta}$$

$$h_{T}(t) = \frac{\frac{1}{\beta}e^{-x/\beta}}{e^{-x/\beta}} = \frac{1}{\beta}$$

b.

$$f_{T}(t) = \frac{\gamma}{\beta}t^{\gamma-1}e^{-t^{\gamma}/\beta}$$

$$F_{T}(t) = P(X^{1/\gamma}\le t) = P(X\le t^{\gamma}) = 1-e^{-t^{\gamma}/\beta}$$

$$h_{T}(t) = \frac{\gamma}{\beta} t^{\gamma-1}$$

c.

$$F_{T}(t) = \frac{1}{1+e^{-(1-\mu)/\beta}}$$

$$f_{T}(t) = \frac{1}{(1+e^{-(1-\mu)/\beta})^{2}} \frac{1}{\beta}e^{-(t-\mu)/\beta}$$

$$= \frac{1}{\beta} F(t) (1-F(t))$$

So $h(t) = \frac{1}{\beta} F(t)$.

3.28

a.

$$f(x\vert\mu,\sigma) = \frac{1}{\sqrt{2\pi}\sigma} \exp\left\{-\frac{(x-\mu)^2}{2\sigma^{2}}\right\}$$

$$= \underbrace{\frac{1}{\sqrt{2\pi}}e^{-\mu^{2}/(2\sigma^{2})}}_{c... ...\underbrace{\frac{\mu}{\sigma^2}}_{w_{2}(\theta)}\right\} \underbrace{1}_{h(x)}$$

b.

$$f(x\vert\alpha,\beta) = \frac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta} 1_{(0,\infty)}(x)$$

$$= \underbrace{\frac{1}{\Gamma(\alpha)\beta^{\alpha}}}_{c(\theta)}... ...{\frac{1}{\beta}}_{w_{2}(\theta)}\right\} \underbrace{1_{(0,\infty)}(x)}_{h(x)}$$

c.

$$f(x\vert\alpha,\beta) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1} 1_{[0,1]}(x)$$

$$= \underbrace{ \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\... ...a)} \underbrace{\log (1-x)}_{t_{2}(x)}\right\} \underbrace{1_{[0,1]}(x)}_{h(x)}$$

d.

$$f(x\vert\lambda) = \frac{\lambda^{x}}{x!}e^{-\lambda} = \underbra... ...\exp\left\{ \underbrace{x}_{t(x)}\underbrace{\log \lambda}_{w(\lambda)}\right\}$$

e.

$$f(x\vert r,p) = \binom{r+x-1}{r}p^{r}(1-p)^{x}$$

$$= \underbrace{\binom{r+x-1}{r}}_{h(x)} \underbrace{p^{r}}_{c(p)} \exp\left\{\underbrace{x}_{t(x)}\underbrace{\log(1-p)}_{w(p)}\right\}$$

3.30

a.

For the binomial, $w(p) = \log(p/(1-p))$, $c(p)=(1-p)^n$, and $t(x)=x$. The variance Var$(t(X))=$Var$(x)$ satisfies

$$(w'(p))^2 (X) = -\frac{d^2}{dp^2}c(p) - w''(p)E[X]$$

Now

$$w'(p) = \frac{1}{p}+\frac{1}{1-p}=\frac{1}{p(1-p)}$$

$$w''(p) = -\frac{1}{p^2}+\frac{1}{(1-p)^2}$$

$$\frac{d^2}{dp^2}c(p) = -\frac{n}{(1-p)^2}$$

So

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

$$= n \left(\frac{1}{1-p}+\frac{1}{p}\right) = \frac{n}{p(1-p)}$$

and thus Var$(X) = n p (1-p)$

b.

For the Beta distribution $t_1(x)=\log x$ and $t_2(x)=\log(1-x)$. The function $f(x)=x$ cannot be expressed as a linear combination of $t_1(x)$ and $t_2(x)$, so the identities in Theorem 3.4.2 cannot be used to find the mean and variance of $X$.

If $X \sim$   Poisson$(\lambda)$ then $t(x)=x$, $w(\lambda) = \log\lambda$, and $c(\lambda = e^{-\lambda}$. So

$$w'(\lambda) = 1/\lambda w''(\lambda) = -1/\lambda^2$$

$$\frac{\partial}{\partial\lambda} \log c(\lambda) = -1 \frac{\partial^2}{\partial\lambda^2} \log c(\lambda) = 0$$

So Theorem 3.4.2 produces the equations

$$E[X/\lambda] = 1$$

$$(X/\lambda) = E[X/\lambda^2]$$

with solutions $E[X] = \lambda$ and Var$(X) = \lambda$.