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.

  • $\begingroup$ where did you get that formula for $\Lambda$? $\endgroup$
    – Taylor
    May 7, 2018 at 2:53
  • 1
    $\begingroup$ I'm not sure why I used the incorrect values there. I edited my expression, so I believe that part is now correct. $\endgroup$
    – antsatsui
    May 7, 2018 at 3:17

1 Answer 1


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


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 $H_1$.

  • $\begingroup$ Forgot to define $f_{\lambda}$: it is the population pmf of course. $\endgroup$ Oct 13, 2018 at 19:28

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