Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1 / 18H (Statistics)

The moment-generating function of $W_i\sim \mathrm{Exp}(1)$ isIndependence makes the moment-generating function of the sum equal the product:This is the moment-generating function of the gamma distribution $\Gamma (n,1)$, and moment-generating functions determine distributions in a neighbourhood of zero. Hence

For $y\ge 0$ and $X\sim U(0,1)$,The distribution function is zero for $y<0$, so this is exactly the distribution function of $\mathrm{Exp}(1)$. Therefore

The Neyman-Pearson lemma states that, for testing one simple hypothesis with density $f_0$ against another with density $f_1$, a size-$\alpha$ test that rejects for the largest values of the likelihood ratio $f_1/f_0$ is most powerful among all tests of size at most $\alpha$, with boundary randomization if needed.

The normalized density isFor $\theta =1$ against $\theta =0$, the sample likelihood ratio isIt is strictly decreasing in $S$. Under $H_0$, parts (a) and (b) giveIf $q_{\alpha }$ is the lower $\alpha$-quantile of this gamma distribution, the Neyman-Pearson lemma gives the most powerful size-$\alpha$ critical region

For any fixed $\theta >0$,is again strictly decreasing in the same statistic $S$. Thus the same critical region is most powerful against every $\theta >0$ and is consequently a uniformly most powerful test of $H_0:\theta =0$ against $H_1:\theta >0$.