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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