= Solution
{c}
The <statistical power> of a test $\varphi$ is $\beta(\theta)=\mathbb E_\theta\varphi(X)$, its rejection <probability> at parameter $\theta$. Its size is $\sup_{\theta\in H_0}\beta(\theta)$; for a simple null this is just $\beta(\theta_0)$.
The <Neyman-Pearson lemma> states that for testing a simple null <probability density function> $f_0$ against a simple alternative <probability density function> $f_1$, a test rejecting where $f_1/f_0>k$, accepting where it is below $k$, and if needed randomizing on equality to attain the desired size is most powerful among all tests of that size. It follows by integrating $(\varphi-\varphi_*)(f_1-kf_0)\leq0$ and using the size constraint.
The proposed <probability density function> is nonnegative, and
$$
\int_0^\infty f(x;\theta)dx
=\frac{\theta^2}{\theta+1}
\left(\frac1{\theta^2}+\frac1\theta\right)=1.
$$
For an independent sample, cancellation of the factors $(x_i+1)$ in the <likelihood ratio> gives
$$
\frac{L(\theta_1)}{L(\theta_0)}
=\left[\frac{\theta_1^2/(\theta_1+1)}
{\theta_0^2/(\theta_0+1)}\right]^n
\exp\left((\theta_0-\theta_1)\sum_i x_i\right).
$$
For $\theta_1<\theta_0$ this is increasing in $S=\sum_iX_i$. The most powerful test therefore has <rejection region>
$$
\boxed{S>c_\alpha,\qquad
\mathbb P_{\theta_0}(S>c_\alpha)=\alpha.}
$$
The sum has a continuous positive <probability density function> on $(0,\infty)$, so no boundary randomization is required. The threshold depends on the null and chosen size, but not on which $\theta_1<\theta_0$ was specified. The same test is therefore \b[a <uniformly most powerful test> for the composite alternative $0<\theta<\theta_0$].
Back to article page
