Let $ \mathbf{X}=\left\{X_{1}, X_{2}, \ldots, X_{n}\right\} $ be an independent random sample from the Poisson distribution with parameter $ \theta>0 $. (i) Find the rejection region of the most powerful test for hypotheses: $$ H_{0}: \theta=1 \text { versus } H_{1}: \theta=2 $$ (ii) Find the critical value such that this test has an exact size 0.05.

I could show that the answer to (i) is $\{\sum_{i=1}^nX_i \ge k\}$, but I'm don't know how to compute the exact value of $k$ in (ii). All I could show is that $2n=\chi^2_{\alpha=0.95,df=2k}$

  • $\begingroup$ The critical value is derived solving in $k$ the following probabilty $$\mathbb{P}\left[\sum_{i=1}^{n}X_i\geq k|\theta=1\right]=0.05$$ thus you have to calculate the probabilities of a poisson $Po(n)$ and obviously in most cases to have "exactly" a size of 5% you must use a randomized test, but anyway you cannot do any calculations if you do not fix a certain $n$ $\endgroup$ – tommik May 7 at 7:54

To get an answer it is necessary to fix a certain $n$. So let's set $n=5$

as per Neyman Pearson's Lemma, the critical region is

$$\mathbb{P}[Y\geq k]=0.05$$

where $Y\sim Po(5)$

It is easy to verify with a calculator (or manually in 5 minutes) that

$$\mathbb{P}[Y\geq 10]=3.18\%$$


$$\mathbb{P}[Y\geq 9]=6.81\%$$

It is evident that there's no way to have a non randomized test which gets exactly a 5% size...thus the test must be randomized in the following way:

  • If the sum of the observations is 10 or higher I reject $H_0$

  • If the sum of the observations is 8 or lower I do not reject $H_0$

  • If the sum of the observations is exactly 9 I toss a fair coin and I reject $H_0$ if the coin shows Head.

this can be formalized as follows:

$$ \psi(y) = \begin{cases} 1, & \text{if $y>9$} \\ 0.5, & \text{if $y=9$} \\ 0, & \text{if $y<9$ } \end{cases}$$

And the total size is

$$\alpha=0.5\times P(Y=9)+P(Y>9)=0.5000\times0.0363+0.0318=0.0500$$

.., as requested


Let's take $n = 20$ for an example. Use $T = \sum_{i=1}^{10} X_i$ as your test statistic, retaining $H_0$ for small $T$ and rejecting $H_0$ for large T.

$T \sim \mathsf{Pois}(\lambda = 10\theta).$ The critical value $c=16$ will have $P(T \ge c\,|\,\lambda=10)= 0.049 \approx 0.05,$ but not larger.

qpois(.95, 10)
[1] 15
1 - ppois(15, 10)
[1] 0.0487404

So (without randomization) a test at exactly level 5% is not available because of the discreteness of Poisson distributions. [If you use a normal approximation, you might fool yourself into thinking you can have a teat at exactly 5%, but that would involve a noninteger, thus nonobtainable, $c.]$

$$P(T \ge c) = P\left(\frac{T-\lambda}{\sqrt{\lambda}} \ge \frac{c-10}{\sqrt{10}}\right)$$ $$\approx P\left(Z < \frac{c-10}{\sqrt{10}} = 1.645\right) = 0.05,$$ so $``c = 15.20.$''

However, @Tommik's (+1) randomization method is the only valid way to get a test at exactly level $\alpha=0.05.$


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