Solutions

5.23

For $0 \le z \le 1$

$$P(Z < z\vert X=x) = 1-(1-z)^{x}$$

$$P(Z<z) = \sum_{i=1}^{\infty} \frac{c}{x!}(1-(1-z)^{x}) = 1-\sum_{i=1}^{\infty}\frac{c}{x!}(1-z)^{x}$$

$$= 1-c\sum_{i=1}^{\infty}\frac{(1-z)^{x}}{x!} = 1-c\left[\sum_{i=0}^{\infty}\frac{(1-z)^{x}}{x!}-1\right]$$

$$= 1-c(e^{1-z}-1)=1-\frac{e^{1-z}-1}{e-1} = \frac{e}{e-1}(1-e^{-z})$$

5.24

Assume, without loss of generality, that $\theta=1$. Then for $0 < u < v < 1$

$$f_{X_{(1)},X_{(n)}}(u,v) = \frac{n!}{0!(n-2)!0!}f(u)f(v)F(u)^{u}(F(v)-F(u))^{n-2}(1-F(v))^{0}$$

$$= n(n-1)f(u)f(v)(F(v)-F(u))^{n-2}$$

$$= n(n-1)(v-u)^{n-2}$$

Let

$$Y = X_{(1)}/X_{(n)}$$

$$Z = X_{(n)}$$

The range of $(Y,Z)$ is $\mathcal{B}=[0,1]\times[0,1]$, the inverse transformation is

$$X_{(1)} = YZ$$

$$X_{(n)} = Z$$

The Jacobian determinant is

$$J(y,z) = \det \begin{pmatrix}z & y 0 & 1 \end{pmatrix} = z$$

So for $0 < z < 1$ and $0 < y < 1$

$$f_{Y,Z}(y,z) = n(n-1)(z-yz)^{n-2}z$$

$$= n z^{n-1} \times (n-1)(1-y)^{n-2}$$

Thus $Y$ and $Z$ are independent.

This proof first removes $\theta$ from consideration since it is just a scale parameter. An alternative approach is to note that $X_{(n)}$ is sufficient, $X_{(1)}/X_{(n)}$ is ancillary, and use Basu's theorem in chapter 6.

5.32

a.

Suppose $f$ is continuous at $a$ and $X_{n} \overset{P}{\rightarrow} a$, $a$ a constant. Fix $\varepsilon > 0$. Then there exists a $\delta > 0$ such that

$$\vert f(x)-f(a)\vert < \varepsilon$$

if $\vert x -a\vert < \delta$. So

$$P(\vert f(X_{n})-f(a)\vert < \varepsilon) \ge P(\vert X_{n}-a\vert < \delta)$$

$$\rightarrow 1$$

So $f(X_{n}) \overset{P}{\rightarrow}f(a)$. The result follows if $f(x) = \sqrt{x}$ or $f(x) = 1/x$ and $a > 0$.

b.

$f(x) = \sigma/\sqrt{x}$ is continuous at $x=\sigma^{2}$ if $\sigma > 0$.

5.40

a.

For any $t$ and any $\varepsilon > 0$, if $X_n > t$ and $\vert X_n-X\vert < \varepsilon$, then $X > t - \varepsilon$. So $X \le t - \varepsilon$ implies that either $X_n \le t$ or $\vert X_n - X\vert \ge \varepsilon$, i.e.

$$\{X \le t - \varepsilon\} \subset \{X_n \le t \} \cup \{\vert X_n - X\vert \ge \varepsilon\}$$

So

$$P(X \le t - \varepsilon) \le P(X_n \le t) + P(\vert X_n - X\vert \ge \varepsilon)$$

or

$$P(X \le t - \varepsilon) - P(\vert X_n - X\vert \ge \varepsilon) \le P(X_n \le t)$$

b.

Similarly (reversing the roles of $X$ and $X_n$ and replacing $t$ by $t + \varepsilon$),

$$P(X_n \le t) \le P(X \le t + \varepsilon) + P(\vert X_n - X\vert \ge \varepsilon)$$

c.

Suppose the CDF of $X$ is continous at $t$. From the previous two parts, since $X_n \rightarrow X$ in probability we have

$$P(X \le t - \varepsilon) \le \liminf_{n \rightarrow \infty} P(X_n... ...t) \le \limsup_{n \rightarrow \infty} P(X_n \le t) \le P(X \le t + \varepsilon)$$

for any $\varepsilon > 0$. Since $t$ is a coninuity point of the distribution of $X$, $\lim_{\varepsilon \downarrow 0}P(X \le t - \varepsilon) = \lim_{\varepsilon \downarrow 0}P(X \le t + \varepsilon) = P(X \le t)$, and therefore

$$P(X \le t) \le \liminf_{n \rightarrow \infty} P(X_n \le t) \le \limsup_{n \rightarrow \infty} P(X_n \le t) \le P(X \le t)$$

So $\lim_{n \rightarrow \infty} P(X_n \le t)$ exists and is equal to $P(X \le t)$. Thus $X_n \rightarrow X$ in distribution.