## Problem 9.10 (2 points)

Let $$Y_1,\dotsc,Y_n$$ be a random sample from a normal distribution with mean $$\mu$$ and variance $$\sigma^2$$. Let $$\hat{\sigma}_2^2 = \frac{1}{2}(Y_1 - Y_2)^2$$. Recall that $$Y_1 - Y_2$$ is normally distributed with mean $$0$$ and variance $$2\sigma^2$$, so $$Z = (Y_1 - Y_2)/(\sigma\sqrt{2})$$ is a standard normal random variable. Therefore $W = \frac{\hat{\sigma}_2^2}{\sigma^2}$ has a $$\chi^2$$-distribution with $$\nu=1$$ degree of freedom. We see that $P(|\hat{\sigma}_2^2 - \sigma^2| > \sigma^2) = P\left(\left|\frac{\hat{\sigma}_2^2}{\sigma^2} - 1\right| > 1 \right) = P(|W - 1| > 1),$ which is a positive constant (about 0.32). Hence $\lim_{n\rightarrow\infty} P(|\hat{\sigma}_2^2 - \sigma^2| > \sigma^2) \ne 0,$ so $$\hat{\sigma}_2^2$$ is not a consistent estimator for $$\sigma^2$$.

## Problem 9.12 (2 points)

Let $$X_1,\dotsc,X_n$$ be a random sample from a normal population with mean $$\mu_1$$ and variance $$\sigma^2$$. Let $$Y_1,\dotsc,Y_n$$ be a random sample from a normal population with mean $$\mu_2$$ and variance $$\sigma^2$$, and that the two samples are independent. Example $$9.3$$ shows that each of \begin{align*} S^2_X &= \frac{1}{n-1}\sum_{i=1}^{n}(X_i - \overline{X})^2 \\ S^2_Y &= \frac{1}{n-1}\sum_{i=1}^{n}(Y_i - \overline{Y})^2 \end{align*}

is a consistent estimator for $$\sigma^2$$. By theorem $$9.2$$(a) $\frac{\sum_{i=1}^{n}(X_i - \overline{X})^2 +\sum_{i=1}^{n}(Y_i - \overline{Y})^2}{2n - 2} = \frac{1}{2}\left( \frac{1}{n-1}\sum_{i=1}^{n}(X_i - \overline{X})^2 +\frac{1}{n-1}\sum_{i=1}^{n}(Y_i - \overline{Y})^2 \right)$ is a consistent estimator for $$\frac{1}{2}(\sigma^2 + \sigma^2) = \sigma^2$$.

## Problem 9.14 (2 points)

Let $$Y \sim \textrm{Binomial}(n,p)$$; then $$E[Y] = np$$, $$V[Y] = np(1-p)$$. Let $$\hat{p}_n = \frac{1}{n}Y$$, then $$E[\hat{p}_n] = p$$ and $$V[\hat{p}_n] = p(1-p)/n$$. Thus $\lim_{n\rightarrow\infty} V[\hat{p}_n] =\lim_{n\rightarrow\infty} \frac{p(1-p)}{n} = 0,$ so, by theorem $$9.1$$, $$\hat{p}_n$$ is a consistent estimator for $$p$$.

## Problem 9.18 (4 points)

Let $$Y_1,Y_2,Y_3,\dotsc$$ be independent standard normal random variables.
1. $$\sum_{i=1}^{n}Y^2_i$$ has a $$\chi^2$$-distribution with $$\nu=n$$ degrees of freedom.
2. Let $$W_n = \frac{1}{n}\sum_{i=1}^{n}Y^2_i$$; then $$E[W_n] = \frac{1}{n}\cdot n = 1$$ and $$V[W_n] = \frac{1}{n^2}\cdot 2n = \frac{2}{n} \rightarrow 0$$ as $$n\rightarrow\infty$$. Thus $$W_n$$ converges in probability to $$1$$ as $$n\rightarrow \infty$$.

## Problem 9.22 (2 points)

Let $$Y_1,\dotsc,Y_n$$ denote a random sample of size $$n$$ from a Pareto distribution. The distribution function for $$Y_{(1)} = \min(Y_1,\dotsc,Y_n)$$ is (see exercise $$6.14$$) $F_{(1)}(y) = \begin{cases} 1 - (\beta / y)^{\alpha\cdot n}, & y > \beta \\ 0, & y \le \beta \end{cases}.$ This allows us to calculate \begin{align*} P(|Y_{(1)} - \beta| \le \epsilon) &= P(\beta - \epsilon \le Y_{(1)} \le \beta + \epsilon) \\ &= F_{(1)}(\beta + \epsilon) - F_{(1)}(\beta - \epsilon) \\ &= 1 - \left(\frac{\beta}{\beta + \epsilon}\right)^{\alpha \cdot n}. \end{align*}

Hence, $\lim_{n\rightarrow \infty} P(|Y_{(1)} - \beta| \le \epsilon) = \lim_{n\rightarrow \infty} \left( 1 - [\beta / (\beta + \epsilon)]^{\alpha \cdot n}\right) = 1.$ Therefore, $$Y_{(1)}$$ converges in probability to $$\beta$$.

## Problem 9.24 (2 points)

Let $$Y_1, \dotsc, Y_n$$ be independent random variables, each with probability density function $f(y) = \begin{cases} 3 y^2, & 0 \le y \le 1 \\ 0, & \textrm{elsewhere} \end{cases}.$ Since this is the density of a Beta distribution with parameters $$\alpha=3$$, $$\beta=1$$, each $$Y_i$$ has mean $$3/4$$ and variance $$3/80$$. The variance of $$\overline{Y}$$ is therefore $$\frac{3}{80 n}\rightarrow 0$$ as $$n\rightarrow \infty$$. Theorem $$9.1$$ implies that $$\overline{Y}$$ converges in probability to $$3/4$$.

## Problem 9.26 (2 points)

Let $$Y_1,\dotsc,Y_n$$ denote a random sample from the probability density function $f(y) = \begin{cases} 2/y^2 , & y \ge 2 \\ 0 , & \textrm{elsewhere} \end{cases}.$ Now, \begin{align*} \lim_{a\rightarrow \infty}\int_2^a y\cdot f(y)\,dy &= \lim_{a\rightarrow \infty}\int_2^a \frac{2}{y}\,dy \\ &= \lim_{a\rightarrow \infty} 2 \ln(y) \Big{\vert}_2^a \\ &= \lim_{a\rightarrow \infty} 2(\ln(a) - \ln(2)) \\ &= \infty, \end{align*}

which shows that $$\int_{-\infty}^{\infty} y\cdot f(y)\,dy$$ diverges. Thus $$E[Y]$$ does not exist. The weak law of large numbers, therefore, does not apply since it assumes a finite expected value.

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