Best Critical Region for Poisson Random Variable

Question

Let $X_1,\ldots, X_{10}$ be a random sample from a Poisson distribution with mean $\theta$. Show that the best critical region against the hypothesis $H_0: \theta = 0.1$ and $H_1: \theta = 0.5$ is $C = \{(x_1,\ldots,x_n)| \sum_{i=1}^{10} x_i \geq3\}$.

My problem is getting the sum to be greater than equal to 3. I get that $\sum_{i=1}^{10} x_i \geq \frac{4-ln(k)}{ln(5)}$ by the Neyman-Pearson theorem, but I don't understand why this implies this is greater than or equal to 3.

Answer 1

I will give you an outline and some relevant computations from R.

Let $S = \sum_{i=1}^{10} X_i.$ Under $H_0,$ we have $S \sim \mathsf{Pois}(\lambda = 1)$ and under $H_1$ we have $S \sim \mathsf{Pois}(\lambda = 1).$

The Neyman-Pearson theorem show that we should reject $H_0$ for large $S.$ Because one would expect larger values of $S$ if $H_1$ is true, it makes sense that the critical region should be of the form $S \ge c,$ but we need to do computations involving Poisson distributions in order to know what significance level $\alpha$ is implied by a specific critical value $c.$ The statement of your problem suggests using $c = 3.$

The significance level is $$\alpha = P(\mathrm{Rej}\,H_0) | H_0\,\mathrm{True}) = P(S \ge 3|\lambda=1)\\ = 1 - P(S \le 2|\lambda=1) \approx 0.08,$$ which agrees with @lonzaleggiera's second comment. Computation in R, where ppois is a Poisson CDF:

1 - ppois(2, 1)
[1] 0.0803014

Then the power $\pi$ of the test with this specified critical region is as follows:

$$\pi = P(\mathrm{Rej}\,H_0) | H_1\,\mathrm{True}) = P(S \ge 3|\lambda=5)\\ = 1 - P(S \le 2|\lambda=5) \approx 0.875.$$

1 - ppois(2, 5)
[1] 0.875348

In the figure below the blue bars correspond to the null distribution $\mathsf{Pois}(1)$ and the brown bars to the alternative distribution $\mathsf{Pois}(5).$ The critical region is to the right of the dotted vertical line. The significance level is given by the sum of the heights of the blue bars in the critical region and the power is given by the sum of the heights of the brown bars.

R code for figure:

s = 0:12;  pdf.0 = dpois(s, 1);  pdf.1 = dpois(s, 5)
hdr = "Null and Alternative Distn's"
plot(s-.05, pdf.0, type="h", col="blue", lwd=3, ylab="PDF", xlab="s", main=hdr)
 lines(s+.05, pdf.1, type="h", col="brown", lwd=3)
 abline(h=0, col="green2")
 abline(v = 2.5, lwd=2, lty="dotted")