# How can I construct an asymptotic confidence interval using a specified pivotal quantity and the score test?

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.

• 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! – Ron Snow Apr 23 at 5:13