Consider $90$ iid geometric random variables having parameter $\theta$ (and thus mean $1/\theta$), and a UMP test of the null hypothesis that $\theta \leq 0.1$ against the alternative that $\theta \gt 0.1$. Give the approximate p-value (based on large sample results) which results for the case of the sum of the observations equal to $723$.

This is a practice problem (not homework). I'm given that the answer should be $0.025$.

I have that if $X_i$ are iid random variables from a full one-parameter exponential family distribution, then for a UMP test of

$$H_0:\theta\leq\theta_0 \text{ vs. } H_1:\theta\gt\theta_0$$

we can use

$$T\left(\vec{X}\right)=\sum_{i=1}^n t(X_i)$$

as a test statistic.

We have that the geometric distribution is a full one-parameter exponential family with

$h(x)=I_{\{1,2,...,\}}(x)$, $c(\theta)=\theta$, $t(x)=x-1$, and $w(\theta)=\text{log}(1-\theta)$

We can use

$$T\left(\vec{X}\right)=\sum_{i=1}^n t(X_i)$$

as a test statistic and reject $H_0$ for sufficiently small values of $T(\vec{X})$ since $w(\theta)$ is decreasing.

Applying the CLT to obtain an approximate p-value, we get




So $$\frac{T\left(\vec{X}\right)-n\cdot\mu_{t,0}}{\sqrt{n}\cdot\sigma_{t,0}}\overset{d}{\longrightarrow} \mathsf N(0,1)$$



Any ideas on where I went wrong?


1 Answer 1


Since $t(x) = x-1$, a sufficient statistic for the data is the sample mean $\bar{x}_n$, which can be used as your test statistic. For the geometric distribution this has moments:

$$\mathbb{E}(\bar{X}_n) = \frac{1}{\theta} \quad \quad \quad \mathbb{V}(\bar{X}_n) = \frac{1-\theta}{n \theta^2}.$$

For large $n$ we can appeal to the central limit theorem to obtain the approximate distribution:

$$\frac{\bar{X}_n - \mathbb{E}(\bar{X}_n)}{\sqrt{\mathbb{V}(\bar{X}_n) }} = \sqrt{n} \cdot\frac{\theta \bar{X}_n - 1}{\sqrt{1-\theta}} \sim \text{N}(0,1).$$

Consider the hypotheses $H_0: \theta \leqslant \theta_0$ and $H_A: \theta > \theta_0$. For these hypotheses, smaller values of the sample mean are more conducive to the alternative hypothesis, and so you have p-value:

$$\begin{equation} \begin{aligned} p(x_1,...,x_n) &= \sup_{\theta \leqslant \theta_0} \mathbb{P}( \bar{X}_n \leqslant \bar{x}_n | \theta) \\[6pt] &= \sup_{\theta \leqslant \theta_0} \mathbb{P} \Bigg( \sqrt{n} \cdot\frac{\theta \bar{X}_n - 1}{\sqrt{1-\theta}} \leqslant \sqrt{n} \cdot\frac{\theta \bar{x}_n - 1}{\sqrt{1-\theta}} \Bigg| \theta \Bigg) \\[6pt] &\approx \sup_{\theta \leqslant \theta_0} \Phi \Bigg( \sqrt{n} \cdot\frac{\theta \bar{x}_n - 1}{\sqrt{1-\theta}} \Bigg) \\[6pt] &= \Phi \Bigg( \sqrt{n} \cdot\frac{\theta_0 \bar{x}_n - 1}{\sqrt{1-\theta_0}} \Bigg) \\[6pt] &= \Phi \Bigg( \frac{1}{\sqrt{n}} \cdot\frac{\theta_0 n \bar{x}_n - n}{\sqrt{1-\theta_0}} \Bigg). \\[6pt] \end{aligned} \end{equation}$$

Your example: In your particular case you have $\theta_0 = 0.1$, $n = 90$ and $\bar{x}_n = 723/90$. So your approximate p-value should be:

$$\begin{equation} \begin{aligned} p &\approx \Phi \Bigg( \frac{1}{\sqrt{90}} \cdot\frac{0.1 \cdot 723 - 90}{\sqrt{1-0.1}} \Bigg) \\[6pt] &= \Phi \Bigg( \frac{1}{\sqrt{90}} \cdot\frac{72.3 - 90}{\sqrt{0.9}} \Bigg) \\[6pt] &= \Phi \Bigg( - \frac{1}{\sqrt{90}} \cdot\frac{17.7}{\sqrt{0.9}} \Bigg) \\[6pt] &= \Phi \Bigg( - \frac{17.7}{\sqrt{81}} \Bigg) \\[6pt] &= \Phi \Bigg( - \frac{17.7}{9} \Bigg) \\[6pt] &= \Phi ( - 1.9 \bar{6} ) \\[6pt] &= 0.02461083. \\[6pt] \end{aligned} \end{equation}$$

In your calculation you made an error in the last line where you substituted $T(\vec{X}) = 723$ instead of $T(\vec{X}) = 723-90 = 633$ (consistent with your definition of $T(\vec{X})$). If you correct this error then you get the same value as in my working.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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