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 $\theta$?


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.

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  • $\begingroup$ 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

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