Uniformly most powerful size $\alpha = .05$ test Suppose $X$ is a single observation from a $\text{Poisson}(\lambda)$ distribution. Determine the uniformly most powerful size $\alpha = .05$ test for the hypothesis test $H_0: \lambda = 5$ and $H_1: \lambda = 10$.
My thoughts: The Neyman-Pearson Lemma tells us the UMP level $\alpha$ test is the likelihood ratio test. The likelihood ratio function is 
$\Lambda = \frac{5^x e^{-5}}{x!}/ \frac{10^x e^{-10}}{x!} = 2^{-x}e^5$. We reject the null hypothesis if $\Lambda < k$ which is equivalent to rejecting if $x > k^*$.
Our hypothesis test will be of the form
$$\phi(x) = 
\begin{cases}
1 \hfill &\text{if $x > k^*$} \\
\gamma \hfill &\text{if $x = k^*$} \\
0 \hfill &\text{otherwise}
\end{cases}
$$
To make a size $\alpha$ test we want to solve for $\gamma$ in the following:
$$\alpha = .05 = \text{E}_{\lambda=5}(\phi(x)) = P_{\lambda=5}(X>k^*) + \gamma P_{\lambda=5}(X=k^*).$$
Can someone offer some confirmation/guidance on this? I am not confident in my method.
 A: In case you are still interested, your work upto this point was correct as far as I can tell.
Suppose $\lambda\in\Theta$ where $\Theta=\mathbb R^{+}$ is the parameter space.
By Neyman-Pearson lemma, a most powerful test of size $\alpha$ for testing $H_0:\lambda=\lambda_0$ against $H_1:\lambda=\lambda_1(\ne\lambda_0)$ is usually defined as
\begin{align}
\phi(x)=\begin{cases}1&,\text{ if }\Lambda(x)>c\\\gamma&,\text{ if }\Lambda(x)=c\\0&,\text{ if }\Lambda(x)<c\end{cases}
\end{align}
, where 
$$\Lambda(x)=\frac{f_{\lambda_1}(x)}{f_{\lambda_0}(x)}$$
and $c\,(>0)$ and $\gamma\in(0,1)$ are so chosen that $$E_{\lambda=\lambda_0}\,\phi(X)=\alpha$$
Therefore in this case,
$$\Lambda(x)=e^{-5}2^x\quad,\,\text{ an increasing function of }x$$
That is, for some positive $k$, $$\Lambda(x)>c\implies x>k$$
So the test $\phi$ is of the form
\begin{align}
\phi(x)=\begin{cases}1&,\text{ if }x>k\\\gamma&,\text{ if }x=k\\0&,\text{ if }x<k\end{cases}
\end{align}
, where $k$ and $\gamma\in(0,1)$ are so chosen that 
\begin{align}
E_{\lambda=5}\,\phi(X)=0.05
\end{align}
That is, $$P(Y>k)+\gamma P(Y=k)=0.05\quad,\, Y\sim \text{Poisson}(5)$$
I used WolframAlpha to get 
$$P(Y>8)\approx 0.068094\qquad\text{ and }\qquad P(Y>9)\approx 0.031838$$
So looks like $k=9$ is the appropriate value subject to the size restriction and corresponding to that, we get $$\gamma\approx 0.501075$$
Finally, the test function is 
\begin{align}
\phi(x)=\begin{cases}1&,\text{ if }x>9\\0.501075&,\text{ if }x=9\\0&,\text{ if }x<9\end{cases}
\end{align}
Since the above test is independent of $\lambda_1=10$, $\phi$ is most powerful against all alternatives $\lambda_1\in \Theta_1$ where $\Theta_1\subset \Theta-\{\lambda_0\}$. Hence $\phi$ is uniformly most powerful of size $\alpha$ for testing $H_0$ versus $H_1$.
