= Solution
{c}
Here the <gamma distribution> uses a rate <statistical parameter> $\theta$, with known shape $\lambda$. Write $S=\sum_iX_i$ and $a=n\lambda$. For each fixed $\theta_1>\theta_0$, the alternative-to-null <likelihood ratio> is
$$
\frac{L(\theta_1)}{L(\theta_0)}
=\left(\frac{\theta_1}{\theta_0}\right)^a
\exp\{-(\theta_1-\theta_0)S\},
$$
strictly decreasing in $S$. The <Neyman-Pearson lemma> therefore gives the same lower-tail <critical region> $S<c_\alpha$ for every larger rate. Since $S$ has a continuous <gamma distribution> under the <null hypothesis>, there is a unique positive $c_\alpha$ satisfying
$$
\mathbb P_{\theta_0}(S<c_\alpha)=\alpha.
$$
Thus \b[reject when $S<c_\alpha$] is a <uniformly most powerful test> of exact <test size> $\alpha$. If $G_a$ denotes the unit-rate <gamma distribution> <cumulative distribution function> with shape $a$, then
$$
c_\alpha=\frac{G_a^{-1}(\alpha)}{\theta_0},\qquad
\pi(\theta)=G_a(\theta c_\alpha).
$$
This is the <uniformly most powerful test for a gamma rate>.
When $\lambda=1/n$, the sum has shape one and hence an <exponential distribution> with rate $\theta$. Therefore
$$
1-e^{-\theta_0c_\alpha}=\alpha,\qquad
\boxed{c_\alpha=-\frac{\log(1-\alpha)}{\theta_0}},\qquad
\boxed{\pi(\theta)=1-(1-\alpha)^{\theta/\theta_0}}.
$$
In particular $\pi(\theta_0)=\alpha$, and the <power function> is strictly increasing, exceeding $\alpha$ for $\theta>\theta_0$ and tending to one as $\theta\to\infty$.
Back to article page
