Solutions

5.30

Let $Y=\overline{X}_{1}-\overline{X}_{2}$. Then

$$E[Y] = 0$$

$$(Y) = 2\frac{\sigma^{2}}{n}$$

So

$$P(\vert Y\vert<\sigma/5) = P\left(\frac{\vert Y\vert}{\sqrt{2\sig... ...sqrt{2}}\right) \approx P\left(\vert Z\vert < \frac{\sqrt{n}}{5\sqrt{2}}\right)$$

Sinze $z_{0.005} = 2.576$,

$$2.576 \approx \frac{\sqrt{n}}{5\sqrt{2}}$$

$$\sqrt{n} \approx \sqrt{2} \times 5 \times 2.576 = 18.215$$

$$n \approx (18.215)^{2} = 331.78$$

So $n = 332$ would do if $Y \sim N(\cdot,\cdot)$. Since this $n$ is quite large, it should probably be close to right.

5.31

Since $z_{0.05} = 1.645$,

$$P\left(\left\vert \frac{\overline{X}-\mu}{\sigma/\sqrt{n}}\right\vert < z_{0.05}\right) \approx 0.9$$

or

$$0.9 \approx P(-0.3 \times 1.645 < \overline{X}-\mu < 0.3 \times 1.645)$$

$$=P(-0.493 < \overline{X}-\mu < 0.493)$$

By Chebychev's inequality,

$$P\left(\left\vert\frac{\overline{X}-\mu}{\sigma/\sqrt{n}}\right\vert \le k\right) \ge 1-\frac{1}{k^{2}}$$

Taking $1/k^{2}=0.1$ means $k^{2}=10$ or $k=\sqrt{10}=3.162$. So

$$0.9 \le P(-0.3 \times 3.162 < \overline{X}-\mu < 0.3 \times 3.162)$$

$$= P(-0.9487 < \overline{X}-\mu < 0.9487)$$

First Problem

a.

A $\chi^{2}_{n}$ random variable $X_{n}$ has the same distribution as $\sum_{i=1}^{n}U_{i}$ with the $U_{i}$ $i.i.d.$ $\chi^{2}_{1}$ random variables. So the central limit theroem states that

$$X_{n} \sim AN(n,2n)$$

as $n \rightarrow \infty$.

b.

Let $V_{n}=\sqrt{X_{n}/n} = f(X_{n}/n)$. Now $X_{n}/n \overset{P}{\rightarrow}1$, $X_{n}/n \sim AN(1, 2/n)$, $f(1)=1$, and $f'(1)=1/2$. So

$$V_{n} \sim AN\left(1,\frac{1}{4}2/n\right) = AN\left(1,\frac{1}{2n}\right)$$

and thus

$$Y_{n} = \sqrt{n}V_{n} \sim AN(\sqrt{n},1/2)$$

c.

Look at graphs of the densities, or at quantile plots.

5.62

a.

The density ratio for a Cauchy envelope is

$$\frac{\frac{1}{\sqrt{2\pi}}e^{-x^2/2}}{\frac{1}{\pi}\frac{1}{1+x^2}} = \sqrt{\frac{\pi}{2}}(1+x^2)e^{-x^2/2}$$

Since this is symmetric about the origin we can maximize for $x \ge 0$. The ratio is maximized at the same point as its logarithm, which is maximized at the value of $u = x^2$ that maximizes $\log(1+u)-u/2$. The derivative of this exmpression with respect to $u$ is $\frac{1}{1+u}-\frac{1}{2}$ with a root at $u = 1$. Since the derivative is decreasing on the nonnegative half line this is a global maximum. So the maximizing value of $x$ is $x=1$ and

$$M = \sqrt{\frac{\pi}{2}} 2 e^{-1/2} = \sqrt{2\pi}e^{-1/2} = 1.520347$$

b.

The density ratio for a double exponential envelope is

$$\frac{\frac{1}{\sqrt{2\pi}}e^{-x^2/2}}{\frac{1}{2}e^{-\vert x\vert}} = \sqrt{\frac{2}{\pi}} e^{\vert x\vert-x/2}$$

For $x \ge 0$ this is also maximized at $x=1$ with

$$M = \sqrt{\frac{2}{\pi}}e^{1/2} = 1.315489$$

c.

The value of $M$ is a little smaller for the double exponential density than for the Cauchy density. This means the expected number of rejections is smaller for the double exponential. Drawing from the double exponential distribution can be done at least as efficiently as drawing from the Caushy distribution, so this suggests that the doulble exponential is a slightly better choice.