Let $(X_1, X_2, X_3)$ be a sample from $Poisson(\lambda)$. Here we test $H_0: \lambda\le 1$ v.s. $H_1: \lambda>1$. Fix $\alpha=0.05$. Find the $\alpha$-level uniformly most powerful test (UMPT).

My work: note that the density function has monotone likelihood ratio on $T=\sum_{i=1}^3 X_i$.

By Karlin-Rubin theorem, the UMPT for test $H_0: \lambda\le 1$ v.s. $H_1: \lambda>1$ is given by \begin{equation} \Phi(X_1,\dots,X_n)= \begin{cases} 1 & \text{if } T>k\\ \gamma & \text{if } T=k\\ 0 & \text{if } T<k \end{cases} \end{equation} where the rejection region is the likelihood ratio $\Lambda>k$, and $k$ and $\gamma$ are decided by $$ \alpha=P_{\lambda_0}(T>k)+\gamma P_{\lambda_0}(T=k). $$

Now my problem is, it seems that I can't solve the gamma. I read this theorem, and some people can't write this gamma, especially for continuous functions. Do we need to consider gamma? I mean the test function is: \begin{equation} \Phi(X_1,\dots,X_n)= \begin{cases} 1 & \text{if } T>k\\ 0 & \text{if } T\le k \end{cases} \end{equation} where the rejection region is the likelihood ratio $\Lambda>k$, and $k$ and $\gamma$ are decided by $$ \alpha=P_{\lambda_0}(T>k). $$

If this one holds, we can solve $k$...

  • 1
    $\begingroup$ $P_{\lambda_0}(T>k)$ is not a continuous function of $k$ for a discrete distribution - it jumps at each integer - so for most $\alpha$ you cannot find a $k$ satisfying $\alpha=P_{\lambda_0}(T>k)$. What did you get for $k$ when you tried to find the critical region in this particular case? $\endgroup$
    – Henry
    Apr 5, 2023 at 7:29

1 Answer 1


An apt quote ($\rm [I]$):

... First $k$ is adjusted so that $P_0(L(X) > k)$ and $P_0(L(X) \geq k)$ bracket $\alpha,$ and then a value $\gamma\in [0, 1]$ is chosen for $\varphi(X)$ when $L(X) = k $ to achieve level $\alpha.$

$T(\mathbf X)\sim \mathrm{Poi}(3).$ A quick look at $\mathtt R:$

> qpois(0.95, 3)
[1] 6
> ppois(6, 3)
[1] 0.9664915
> ppois(7, 3)
[1] 0.9880955
> ppois(5, 3)
[1] 0.916082

Now can you choose a particular $k$ from above and as per the quote, then calculate the necessary $\gamma$?


$\rm [I]$ Theoretical Statistics: Topics for a Core Course, Robert W. Keener, Springer Science$+$Business Media, $2010,$ p. $224.$

  • $\begingroup$ So the value of $\gamma$ and $K$ are not unique? $\endgroup$
    – Hermi
    Apr 5, 2023 at 12:05
  • $\begingroup$ MP tests agree with each other except on a set of measure zero. Re-read the quote. You won't find multiple $k$s. $\endgroup$ Apr 5, 2023 at 15:21
  • 1
    $\begingroup$ straight to the point (+1) also for the reference to one of my preferred books! $\endgroup$
    – utobi
    Apr 6, 2023 at 19:59

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