Stat 5102 (Geyer) Midterm 1

Problem 1

The basic fact this problem uses is

$$\frac{X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n - \mu}{S_n / \sqrt{n}} \sim t(n - 1)$$

(Corollary 7.25 in the notes), which, since $\mu = 0$, specializes to

$$\frac{X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n}{S_n / \sqrt{n}} \sim t(n - 1)$$ (1)

in this problem. This is the only exact result we have involving both $X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n$ and $S_n$, so nothing else is of any use.

To use (1), we must put the event of interest $X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n < S_n$ in a form related to the left hand side of (1). Clearly, this is equivalent to

$$\frac{X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n}{S_n / \sqrt{n}} < \sqrt{n} = 3$$

So the probability we need to find $P(Y < 3)$ where $Y$ is a $t(n - 1)$ random variable ($n - 1 = 8$ degrees of freedom).

From Table IIIa in Lindgren $P(Y > 3) = 0.009$, so $P(Y < 3) = 1 - 0.009 = 0.991$.

Problem 2

To use the method of moments, we first need to find some moments. Since this is not a ``brand name'' distribution, we must integrate to find the moments. The obvious moment to try first is the first moment (the mean)

$$\begin{split} \mu & = \int_0^1 x f_\beta(x) \, d x \\ ... ...ht]_0^1 \\ & = \frac{1 + 2 \beta}{3 + 3 \beta} \end{split}$$

Solving for $\beta$ as a function of $\mu$, we get

$$\beta = \frac{3 \mu - 1}{2 - 3 \mu}$$

(the numerator and denominator are both positive because $1 / 3 < \mu < 2 / 3$.)

Either way, we get a method of moments estimator by plugging in $X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n$ for $\mu$

$$\hat{\beta}_n = \frac{3 X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n - 1}{2 - 3 X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n}$$

Problem 3

(a)

The asymptotic distribution of $X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n$ is, as usual, by the CLT,

$$X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n \approx \NormalDis\left(\mu, \frac{\sigma^2}{n}\right).$$

Plugging in $\sigma^2 = \mu$ gives

$$X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n \approx \NormalDis\left(\mu, \frac{\mu}{n}\right).$$

(b)

The asymptotic distribution of $V_n$ is, by Corollary 7.17 in the notes,

$$V_n \approx \NormalDis\biggl(\mu_2, \frac{\mu_4 - \mu_2^2}{n}\biggr).$$

where $\mu_2 = \sigma^2 = \mu$ and $\mu_4 = \mu + 3 \mu^2$ are given in the problem statement. Plugging these in gives

$$V_n \approx \NormalDis\biggl(\mu, \frac{\mu + 2 \mu^2}{n}\biggr).$$

(c)

The ARE is the ratio of the asymptotic variances, either $1 + 2 \mu$ or the the reciprocal $1 / (1 + 2 \mu)$, depending on which way you write it.

(d)

The better estimator is the one with the smaller asymptotic variance, in this case $X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n$.

Problem 4

This is a problem for the delta method. We know from the properties of the exponential distribution

$$E(X_i) = \frac{1}{\lambda}$$

and

$$\var(X_i) = \frac{1}{\lambda^2}$$

Hence the CLT says in this case

$$X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n \approx \NormalDis\left(\frac{1}{\lambda}, \frac{1}{n \lambda^2}\right)$$

For any differentiable function $g$, the delta method says

$$g(X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n) \approx \Nor... ...\lambda}\Bigr), g'\Bigl(\frac{1}{\lambda}\Bigr)^2 \frac{1}{n \lambda^2}\right)$$

The $g$ such that $W_n = g(X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n)$ is

$$g(x) = \frac{x}{1 + x}$$ (2)

which has derivative

$$g'(x) = \frac{1}{(1 + x)^2}$$ (3)

so

$$g\Bigl(\frac{1}{\lambda}\Bigr) = \frac{1}{1 + \lambda}$$

and

$$g'\Bigl(\frac{1}{\lambda}\Bigr) = \frac{\lambda^2}{(1 + \lambda)^2}$$

and

$$W_n \approx \NormalDis\left(\frac{1}{1 + \lambda}, \frac{\lambda^2}{n (1 + \lambda)^4} \right)$$

Problem 5

There are several different ways to proceed here.

Using a Confidence Interval for the Mean

The mean is $\mu = 1 / p$. Thus we could just get a confidence interval for $\mu$ and take reciprocals of the endpoints to get a confidence interval for $p$.

A confidence interval for $\mu$ can be found using Theorem 9.8 in the notes. From Section B.1.8 of the notes

$$\var(X_i) = \frac{1 - p}{p^2} = \mu (\mu - 1)$$

so by the LLN and the continuous mapping theorem

$$S_n = \sqrt{X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n (X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n - 1)}$$

is a consistent estimator of the population standard deviation $\sigma$ needed for the theorem. The theorem gives

$$X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n \pm 1.96 \sqrt{... ...ntom{\text{X}}}_n (X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n - 1)}{n}}$$

as an asymptotic 95% confidence interval for $\mu$. So

$$\frac{1}{X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n + 1.96... ...tom{\text{X}}}_n (X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n - 1)}{n}}}$$

is an asymptotic 95% confidence interval for $p = 1 / \mu$. Plugging in the numbers gives

$$0.180 < p < 0.256$$

Using the Delta Method

The obvious point estimator for $p$ is

$$\hat{p}_n = \frac{1}{X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n}$$

The CLT says

$$X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n \approx \NormalDis\left( \frac{1}{p}, \frac{(1 - p)}{n p^2} \right)$$

Applying the delta method with the transformation

$$g(u) = \frac{1}{u}$$

with derivative

$$g'(u) = - \frac{1}{u^2}$$

gives

$$g\Bigl(\frac{1}{p}\Bigr) = p$$

and

$$g'\Bigl(\frac{1}{p}\Bigr) = - p^2$$

and

$$\hat{p}_n \approx \NormalDis\left( p, \frac{p^2 (1 - p)}{n} \right)$$

which gives an asymptotic 95% confidence interval

$$\hat{p}_n \pm 1.96 \sqrt{\frac{\hat{p}_n^2 (1 - \hat{p}_n)}{n}}$$

Plugging in the numbers gives

$$0.2114 \pm 0.03680$$

or

$$0.1746 < p < 0.2482$$

which is pretty close to the other interval.

Solving Quadratic Inequalities

The really hard way to do this problem is to start with

$$X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n \approx \NormalDis\left(\frac{1}{p}, \frac{1 - p}{n p^2}\right)$$

and standardize giving the asymptotically standard normal quantity

$$\frac{X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n - \frac{1}{p}}{\sqrt{\frac{1 - p}{n p^2}}}$$

from which we conclude that the set of $p$ such that

$$\left\lvert \frac{X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n - \frac{1}{p}}{\sqrt{\frac{1 - p}{n p^2}}} \right\rvert < 1.96$$

is an asymptotic 95% confidence interval for $p$. It turns out this is solvable, equivalent to

$$\begin{split} 1.96^2 & > \left\lvert \frac{X{\mkern -13.5 ... ...mu}\overline{\phantom{\text{X}}}_n p - 1)^2}{1 - p} \end{split}$$

Or, writing $z = 1.96$, we see the confidence interval has endpoints satisfying the quadratic equation

$$(1 - p) z^2 = n (X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n p - 1)^2$$

which has roots

$$\frac{2 n X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n - z^2... ...antom{\text{X}}}_n}} {2 n X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n^2}$$

or

$$0.17375 < p < 0.247366$$