 Solution

A simple hypothesis specifies the entire probability distribution. A critical region $C$ is the set of observations for which the null hypothesis is rejected. The size of a statistical test is ${\sup }_{\theta \in H_0}{\mathrm{Pr}}_{\theta }(X\in C)$, which reduces to ${\mathrm{Pr}}_0(C)$ for a simple hypothesis. The power function is ${\mathrm{Pr}}_{\theta }(C)$ as a function of the true parameter or distribution. At a specified alternative, the Type II error probability is ${\mathrm{Pr}}_{\theta }(C^c)=1-{\mathrm{Pr}}_{\theta }(C)$, the chance of failing to reject a false null hypothesis. The Type I error probability is the corresponding rejection probability under the null hypothesis.

The Neyman-Pearson lemma says that, for two simple hypotheses, a test rejecting where $f_1>k{f}_0$, accepting where $f_1<k{f}_0$, and if necessary randomizing on equality to achieve prescribed size $\alpha$, has maximal statistical power among tests of size at most $\alpha$. Conversely a most powerful test can be taken in this likelihood ratio form, up to equality sets and null sets.

Here the likelihood ratio for $x\ne 0$ isIts value is infinite at $0$, which has probability zero under both continuous distributions. Large likelihood ratios correspond to small $|x|$. Thus the best critical region is $|X|<c_{\alpha }$, whereThere is no boundary randomization to worry about. The complete answer is

For the minimum sum of errors in a simple hypothesis test, write the error sum as ${\int }_C{f}_0+{\int }_{C^c}{f}_1$. At each observation its smaller contribution is obtained by rejecting exactly where $f_1>f_0$. This proves optimality over all tests, including randomized ones, rather than just optimizing within an assumed family. Here it gives $c_{\ast }=\sqrt{2/\pi }$, and thereforeThe minimum error sum is $1-e^{-1/\pi }+2[1-\Phi (\sqrt{2/\pi })]$. This criterion weights the two error probabilities equally; it does not impose a conventional $5\%$ significance level.