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Let $p$ be the row player's mixed strategy and $q$ the column player's mixed strategy. The fourth row is strictly dominated by the first. A useful candidate mixes the first two rows so that their first and third column payoffs agree. If the first row has probability $t$, these payoffs are $7t-2$ and $2-7t$, giving $t=2/7$ and common value zero.

To prove that these are optimal mixed strategies, use the two explicit mixed strategiesTheir payoff vectors satisfyThus the row player guarantees a payoff at least zero against every column mixed strategy, and the column player guarantees a payoff at most zero against every row mixed strategy. This matching pair proves that the value of the zero-sum game is zero.

These optimal mixed strategies are also unique. Since $Aq$ is strictly negative in rows three and four, any optimal row strategy gives them probability zero. Against columns one and three it must then satisfy both $7t-2\ge 0$ and $2-7t\ge 0$, so $t=2/7$. Similarly, $p^{T}A$ is strictly positive in column two, forcing an optimal column strategy to avoid that column; its constraints against rows one and two then force equal weights on columns one and three.

Let $S={\sum }_{i=1}^n{X}_i$. The likelihood ratio of rate two to rate one is $2^n{e}^{-S}$, which decreases strictly with $S$. By the Neyman-Pearson lemma, the most powerful test rejects for small values of the sum:Under the null, $S$ has the gamma distribution with shape $n$ and rate one. Thus $c_{\alpha }$ is its $\alpha$-quantile, equivalently half the $\alpha$-quantile of the chi-squared distribution with $2n$ degrees of freedom. The distribution is continuous, so boundary randomization is unnecessary for $0<\alpha <1$.

For every fixed alternative $\lambda >1$, the likelihood ratio against rate one is ${\lambda }^n{e}^{-(\lambda -1)S}$, again strictly decreasing. Therefore the same region is most powerful against every such alternative. Moreover, if $G$ has the shape-$n$, rate-one gamma distribution, then $S$ at rate $\lambda$ has the same distribution as $G/\lambda$. It follows that ${\mathbb{P}}_{\lambda }(S<c_{\alpha })=\mathbb{P}(G<\lambda c_{\alpha })$ increases with $\lambda$, so the maximum rejection probability over $\lambda \le 1$ is $\alpha$ at one. Any competing size-$\alpha$ test for the composite null is also of level at most $\alpha$ at rate one, and is bounded in power by the Neyman-Pearson lemma for each alternative. The test is uniformly most powerful for the composite hypotheses as well. This is the uniformly most powerful test for an exponential rate based on the sample sum.