Problem 9.2 (4 points)

Let $$Y_1,\dotsc,Y_n$$ denote a random sample from a population with mean $$\mu$$ and variance $$\sigma^2$$. Let \begin{align*} \hat{\mu}_1 &= \frac{1}{2}(Y_1+Y_2), \\ \hat{\mu}_2 &= \frac{1}{4}Y_1+\frac{Y_2+\dotsb+Y_{n-1}}{2(n-2)}+\frac{1}{4}Y_n, \\ \hat{\mu}_3 &= \overline{Y}. \end{align*}
1. We will calculate the expected values of $$\hat{\mu}_1$$, $$\hat{\mu}_2$$, and $$\hat{\mu}_3$$. \begin{align*} E[\hat{\mu}_1] &= \frac{1}{2}(E[Y_1] + E[Y_2]) = \frac{1}{2}(\mu+\mu)=\mu. \\ E[\hat{\mu}_2] &= \frac{1}{4}E[Y_1]+\frac{1}{2(n-2)}(E[Y_2]+\dotsb+E[Y_{n-1}])+\frac{1}{4}E[Y_n] \\ &= \frac{1}{4}\mu+\frac{1}{2(n-2)}\left(\sum_{i=2}^{n-1}\mu\right)+\frac{1}{4}\mu \\ &= \frac{1}{4}\mu+\frac{1}{2}\mu+\frac{1}{4}\mu \\ &= \mu. \\ E[\hat{\mu}_3] &= E[\overline{Y}] = \mu. \end{align*} We see that each of $$\hat{\mu}_1$$, $$\hat{\mu}_2$$, $$\hat{\mu}_3$$ is an unbiased estimator for $$\mu$$.
2. We will calculate the variances of $$\hat{\mu}_1$$, $$\hat{\mu}_2$$, and $$\hat{\mu}_3$$. \begin{align*} V[\hat{\mu}_1] &= \frac{1}{4}(V[Y_1] + V[Y_2]) = \frac{1}{4}(\sigma^2+\sigma^2)=\frac{1}{2}\sigma^2. \\ V[\hat{\mu}_2] &= \frac{1}{16}E[Y_1]+\frac{1}{4(n-2)^2}(V[Y_2]+\dotsb+V[Y_{n-1}])+\frac{1}{16}V[Y_n] \\ &= \frac{1}{16}\sigma^2+\frac{1}{4(n-2)^2}\left(\sum_{i=2}^{n-1}\sigma^2\right)+\frac{1}{16}\sigma^2 \\ &= \frac{1}{16}\sigma^2+\frac{1}{4(n-2)}\sigma^2+\frac{1}{16}\sigma^2 \\ &= \frac{n}{8(n-2)} \sigma^2. \\ E[\hat{\mu}_3] &= E[\overline{Y}] = \frac{1}{n} \sigma^2. \end{align*} We calculate the relative efficiencies \begin{align*} \mathrm{eff}(\hat{\mu}_3,\hat{\mu}_1) &= \frac{V[\hat{\mu}_1]}{V[\hat{\mu}_3]} = \frac{1}{2} n. \\ \mathrm{eff}(\hat{\mu}_3,\hat{\mu}_2) &= \frac{V[\hat{\mu}_2]}{V[\hat{\mu}_3]} = \frac{n^2}{8(n-2)}. \end{align*}

Problem 9.4 (2 points)

Let $$Y_1,\dotsc,Y_n$$ denote a random sample of size $$n$$ from a uniform distribution on the interval $$(0 , \theta)$$. Let $$Y_{(1)} = \min(Y_1,\dotsc,Y_n)$$ and $$Y_{(n)} = \max(Y_1,\dotsc,Y_n)$$. Exercise 8.14 implies that $$\hat{\theta}_1 = (n+1)Y_{(1)}$$ is an unbiased estimator for $$\theta$$, and example 9.1 shows that $$\hat{\theta}_2 = \frac{n+1}{n}Y_{(n)}$$ is an unbiased estimator for $$\theta$$ with $V[\hat{\theta}_2] = \frac{\theta^2}{n(n+2)}.$ Recall that the density for $$Y_{(1)}$$ is $g_{(1)}(y) = \frac{n}{\theta}\left(1 - \frac{y}{\theta}\right)^{n-1},\quad 0 \le y \le \theta,$ so if we let $$u = y/\theta$$ then $$du = \frac{1}{\theta}dy$$ and $g_{(1)}(y)\,dy = n(1 - u)^{n-1}\,du = \frac{\Gamma(n+1)}{\Gamma(1)\Gamma(n)}u^{1-1}(1-u)^{n-1}\,du.$ Therefore, $$U = \frac{1}{\theta}Y_{(1)}$$ has a Beta distribution with parameters $$\alpha=1$$, $$\beta=n$$, and hence \begin{align*} V[\hat{\theta}_1] &= V[(n+1)Y_{(1)}] = V[(n+1)\theta U] \\ &= (n+1)^2 \theta^2 V[U] \\ &= (n+1)^2 \theta^2 \frac{n}{(n+1)^2 (n+2)} \\ &= \frac{n}{n+2} \theta^2. \end{align*}

From this we calculate the relative efficiency $\textrm{eff}(\hat{\theta}_1,\hat{\theta}_2) = \frac{V[\hat{\theta}_2]}{V[\hat{\theta}_1]} = \frac{\theta^2}{n (n+2)}\cdot \frac{(n+2)}{n \theta^2} = \frac{1}{n^2}.$

Problem 9.6 (2 points)

Suppose $$Y_1,\dotsc,Y_n$$ denotes a random sample of size $$n$$ from a Poisson distribution with parameter $$\lambda$$. Let $$\hat{\lambda}_1 = \frac{1}{2}(Y_1 + Y_2)$$, $$\hat{\lambda}_2 = \overline{Y}$$. Then \begin{align*} V[\hat{\lambda}_1] &= \frac{1}{4}(V[Y_1] + V[Y_2]) = \frac{1}{2}\lambda. \\ V[\hat{\lambda}_2] &= V[\overline{Y}] = \frac{1}{n}\lambda. \\ \textrm{eff}(\hat{\lambda}_1,\hat{\lambda}_2) &= \frac{V[\hat{\lambda}_2]}{V[\hat{\lambda}_1]} = \frac{\lambda}{n}\cdot \frac{2}{\lambda} = \frac{2}{n}. \end{align*}

Problem 9.8 (4 points)

1.   \begin{align*} Y &\sim N(\mu , \sigma^2) \\ f(y) &= \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\frac{(y-\mu)^2}{\sigma^2}} \\ \ln f(y) &= -\frac{1}{2}(y-\mu)^2 / \sigma^2 - \ln(\sigma \sqrt{2\pi}) \\ -\frac{\partial^2 \ln f}{\partial \mu^2} &= \frac{1}{\sigma^2} \\ E\left[-\frac{\partial^2 \ln f}{\partial \mu^2}(Y)\right] &= \frac{1}{\sigma^2}. \end{align*} Therefore, $I(\mu) = \frac{1}{n/\sigma^2} = \frac{\sigma^2}{n} = V[\overline{Y}],$ which shows that $$\overline{Y}$$ is an efficient estimator for $$\mu$$.
2.   \begin{align*} Y &\sim \textrm{Poisson}(\lambda). \\ p(y) &= e^{-\lambda}\lambda^y / y!, y=0,1,2,\dotsc \\ \ln p(y) &= -\lambda + y\ln \lambda - \ln(y!) \\ -\frac{\partial^2 \ln p}{\partial \lambda^2} &= \frac{y}{\lambda^2} \\ E\left[-\frac{\partial^2 \ln p}{\partial \lambda^2}(Y)\right] &= E[Y/\lambda^2] = 1/\lambda. \end{align*} Therefore, $I(\lambda) = \frac{1}{n/\lambda} = \frac{\lambda}{n} = V[\overline{Y}],$ which shows that $$\overline{Y}$$ is an efficient estimator for $$\lambda$$.
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