How to determine a sufficient statistic for a Poisson sample and show that it has a monotone likelihood ratio.

Question

The following question is a last year's Statistics exam question I tried to solve (without any luck). Any help would be grateful. Thanks in advance.

An Atomic Energy Agency is worried that a particular nuclear plant has leaked radio-active material. They do $5$ independent Geiger counter measurements in the direct neighbourhood of the reactor. They find the following measurements (per unit time):

observation i: 1 2 3 4 5

count $x_i$ : 1 2 6 2 7

(I did not know how to implement this into a tabular)

The natural background radiation has an average of $λ = 2$ (per unit time). The agency would only be worried if the radiation rate would be in the order of $λ = 5$.

They therefore decide to test: $H_0 : λ ≤ 2$ versus : $H_1 : λ > 2$

They want to device the optimal test to see if there there is any reason for alarm. Assuming that the data are realizations of a sample from a Poisson distribution:

$X_1, ..., X_5 ∼ POI(λ)$

with density: $f(x) = e^{-λ}\frac{λ^{x}}{x!}$

I have two questions I need some help with:

1. Determine a sufficient statistic for the Possion sample and show that it has a monotone likelihood ratio.

2. Derive the uniform most powerful test of level $α = 0.0487$ for the test problem.

Because we have a Poisson distribution, I know that we can use: $\sum_{i = 1}^{5}X_i \sim Poi(5λ)$

For the first question, my attempt:

$p(x_1,...,x_5|λ) = \prod_{i = 1}^{5} e^{-5λ}\frac{λ^{x_1 + x_2 + x_3 + x_4 + x_5}}{x_1!x_2!x_3!x_4!x_5!} = h(x_1 +...+ x_5|λ) * g(x_1,x_2,x_3,x_4,x_5) $

$h(x_1 +...+ x_5|λ) = e^{-5λ}λ^{x_1 + x_2 + x_3 + x_4 + x_5} $

$g(x_1,x_2,x_3,x_4,x_5) = \frac{1}{x_1!x_2!x_3!x_4!x_5!}$

It follows by the factorization theorem that $T(X_1, X_2, X_3,X_4,X_5) = X_1+X_2+X_3+X_4+X_5$ is sufficient statistic.

Not sure how to construct a proof to show it has a monotone likelihood ratio.

Answer 1

I will show the test and its result, leaving it to you to justify that it is a LR test based on the sufficient statistic.

The sum $T$ of five readings is 18. Under $H_0: \lambda_T = 5(2) = 10,$ one has $P(T \ge 16) = 0.0487.$ So the critical value for a right-tailed test at the 5% level (or below) is $c=16,$ and $H_0$ is rejected. The P-value is

$$P(T \ge 18\,|\,\lambda_T=10) = 1 - P(T \le 17\,|\,\lambda_T = 10) = 0.014.$$

Computations in R, where ppois denotes Poisson CDF are shown below. For such small $\lambda$ it would not be appropriate to use a normal approximation.

1 - ppois(15,10)
[1] 0.0487404
1 - ppois(17,10)
[1] 0.01427761

Answer 2

Show it has a monotone LR

Let's set $\theta_1 < \theta_2$

The Likelihood Ratio (LR) is the following

$$\frac{L(\theta_1;\mathbf{x})}{L(\theta_2;\mathbf{x})}=\frac{e^{-n\theta_1}\theta_1^{\Sigma x}}{e^{-n\theta_2}\theta_2^{\Sigma x}}=e^{n(\theta_2-\theta_1)}(\frac{\theta_1}{\theta_2})^{\Sigma x}$$

that is a monotone decreasing function in $\Sigma x$