 Solution

The statistical power of a test $\phi$ is $\beta (\theta )={\mathbb{E}}_{\theta }\phi (X)$, 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 $(\phi -{\phi }_{\ast })(f_1-k{f}_0)\le 0$ and using the size constraint.

The proposed probability density function is nonnegative, andFor an independent sample, cancellation of the factors $(x_i+1)$ in the likelihood ratio givesFor ${\theta }_1<{\theta }_0$ this is increasing in $S={\sum }_i{X}_i$. The most powerful test therefore has rejection regionThe 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 a uniformly most powerful test for the composite alternative $0<\theta <{\theta }_0$.