3
$\begingroup$

Let there be a random sample $X_1,...,X_n \sim Poison(\theta)$, where $\theta>0$ is unknown. Show that $P(\mathbf{X},\theta)=\frac{\bar{X}-\theta}{\sqrt{\bar{X}/n}}$ is asymptotically pivotal, then construct as asymptotic $1-\alpha$ confidence interval for $\theta$. Also, construct an asymptotic $1-\alpha$ confidence interval for $\theta$ by inversion of the acceptance region provided by the score test.

my work:

We know that $\bar{X} \sim AN(\theta,\frac{\theta}{n})$.

$P(\mathbf{X},\theta)=\frac{\sqrt{n}(\bar{X}-\theta)/\sqrt{\theta}}{\sqrt{\bar{X}/\theta}}$, where $\frac{\sqrt{n}(\bar{X}-\theta)}{\sqrt{\theta}} \sim AN(0,1)$ and $\bar{X}/\theta \sim AN(0,\frac{1}{n})$.

However, how can I find the distribution of $P(\mathbf{X},\theta)$ given that I know this. I'm not sure what the asymptotic distribution is of the denominator.

Regarding the score test method, I have the following:

We reject $H_0:\theta=\theta_0$ in favor of $H_1:\theta \ne \theta_0$ when $\frac{S^2(\theta_0)}{ni(\theta_0)}>\chi^2_{1;\alpha}$.

We find $i(\theta)=-E[\frac{\partial}{\partial \theta}(-1+\frac{x}{\theta})]=-E[-\frac{x}{\theta^2}]=\frac{1}{\theta}$, since $E(X)=\theta$.

$S(\theta)=\frac{\partial}{\partial \theta} (-n\theta +ln(\theta)\sum x_i -\sum ln(x_i!))=-n + \frac{\sum x_i}{\theta}$.

Thus, we have $\frac{(-n + \frac{\sum x_i}{\theta_0})^2}{\frac{n}{\theta_0}}=\frac{\theta_0n^2-2n\sum x_i +(\sum x_i)^2/\theta_0}{n}>\chi^2_{1;\alpha}$ as our rejection region.

Our acceptance region is $\theta_0^2n^2-2\theta_0n\sum x_i +(\sum x_i)^2 \le \chi^2_{1;\alpha}$. Solving for $\theta_0$, I get $\theta_0=\frac{\sum x_i}{n}$. Where do I go from here to determine the asymptotic $1-\alpha$ confidence interval for $\theta$?

$\endgroup$
2
$\begingroup$

It is unclear whether you want to base your confidence interval on the initial normal approximation or the normal approximation to the score function. I am going to assume the former. You have already found the asymptotic distribution, which can be expressed as:

$$\frac{(\bar{X}-\theta)^2}{\bar{X} / n} \overset{\text{Approx}}{\sim} \text{ChiSq}(1).$$

Take $\chi_{1-\alpha}^2$ to be the critical point of this distribution with upper tail area $\alpha$. Using the polynomial roots derived below, you have:

$$\begin{aligned} 1-\alpha &\approx \mathbb{P} \Bigg( \frac{(\bar{X}-\theta)^2}{\bar{X} / n} \leqslant \chi_{1-\alpha}^2 \Bigg) \\[6pt] &= \mathbb{P} \Bigg( (\bar{X}-\theta)^2 \leqslant \frac{\chi_{1-\alpha}^2}{n} \bar{X} \Bigg) \\[6pt] &= \mathbb{P} \Bigg( \theta^2 - 2 \bar{X} \theta + \bar{X}^2 \leqslant \frac{\chi_{1-\alpha}^2}{n} \bar{X} \Bigg) \\[6pt] &= \mathbb{P} \Bigg( \theta^2 - 2 \bar{X} \theta + \Big( \bar{X} - \frac{\chi_{1-\alpha}^2}{n} \Big) \bar{X} \leqslant 0 \Bigg) \\[6pt] &= \mathbb{P} \Bigg( (\theta - r_1(\bar{X})) (\theta - r_2(\bar{X})) \leqslant 0 \Bigg) \\[6pt] &= \mathbb{P} \Bigg( r_1(\bar{X})^+ \leqslant \theta \leqslant r_2(\bar{X}) \Bigg). \\[6pt] \end{aligned}$$

(Note that we have used the notation for the positive part of the bound on the lower bound; this holds because $\theta>0$.) Thus, substituting the observed data, we obtain the confidence interval:

$$\text{CI}_\theta (1-\alpha) = \Big[ r_1(\bar{x})^+, r_2(\bar{x}) \Big].$$

Note that this is not an especially good confidence interval, since it involves truncating the lower boundary to zero when $\alpha$ is low. Nevertheless, it should serve reasonably well when $n$ is large.


Deriving the polynomoial roots: Define the polynomial:

$$P(\theta, \bar{x}) \equiv \theta^2 - 2 \bar{x} \theta + \Big( \bar{x} - \frac{\chi_{1-\alpha}^2}{n} \Big) \bar{x}.$$

Using quadratic formula, this polynomial has roots:

$$\begin{aligned} r(\bar{x}) &= \frac{1}{2} \Bigg[ 2 \bar{x} \pm \sqrt{4 \bar{x}^2 - 4 \Big( \bar{x} - \frac{\chi_{1-\alpha}^2}{n} \Big) \bar{x} } \Bigg] \\[6pt] &= \frac{1}{2} \Bigg[ 2 \bar{x} \pm \sqrt{ 4 \cdot \frac{\chi_{1-\alpha}^2}{n} \bar{x}} \Bigg] \\[6pt] &= \bar{x} \pm \sqrt{ \frac{\chi_{1-\alpha}^2 \bar{x}}{n}}, \\[6pt] \end{aligned}$$

which we denote seperately as:

$$r_1(\bar{x}) = \bar{x} - \sqrt{ \frac{\chi_{1-\alpha}^2 \bar{x}}{n}} \quad \quad \quad \quad \quad r_2(\bar{x}) = \bar{x} + \sqrt{ \frac{\chi_{1-\alpha}^2 \bar{x}}{n}}.$$

For $\chi_{1-\alpha}^2 \leqslant \dot{x}$ both of these roots are non-negative. When the critical point is above this value (which happens for small values of $\alpha$) the lower root goes below zero and so the confidence interval may not have the approximate coverage probability shown.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Your solution is much more elegant than what I found. In case someone stumbles upon this later: I was able to find the $1-\alpha$ confidence interval by inverting the acceptance region given by the Score Test, where I had to solve for a much more complicated polynomial than what is provided in this answer. I also used the pivotal quantity in the post, since $Q(\mathbf{X},\theta)$ goes in distribution to a standard normal. From there, it's evident how to use the pivotal quantity to obtain the confidence interval. You can show that $Q \sim AN(0,1)$ by using Slutsky's Theorem. Thanks, Ben! $\endgroup$ – Ron Snow Apr 23 at 5:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.