# Neyman-Pearson Lemma for the exponential distribution

I have the following question for my homework:

Suppose X~exp($\theta$).

We want to test $H_0: \theta=1 vs. H_a:\theta=2$, based on a sample of size 2 - ${X_1,X_2}.$

a. Obtain the most powerful test (MPT) at a significance level $\alpha=.01$

What I am confused about is how to go about next steps.

Since I have $H_0: \theta=1 vs. H_a:\theta=2$, should I just plug those in for $\lambda$ to get my likelihood function to be:

$\frac{L_1}{L_0}=\frac{\Pi2e^{-2x}}{\Pi e^{-x}}$

Then I end up with:

$\frac{L_1}{L_0}=\frac{2e^{-n2\Sigma x_i}}{e^{-n\Sigma x_i}}$

$\frac{L_1}{L_0}=\frac{2e^{-2\Sigma x_i}}{e^{-\Sigma x_i}}$

$\frac{L_1}{L_0}=2e^{-\Sigma x_i}$

I am not sure where to go from here or if I am doing this right.

Could someone please let me know whenever they have the chance?

Thanks!

$$Y = \sum_{i=1}^n X_i$$
$$Y \leq c$$
and from the additivity property of the Gamma distribution, the critical value corresponding to a $0.01$ test may be obtained, under $H_0$. This constant turns out to equal $6.63$ for $n=2$.
• Shouldn't the rejection region be $Y>c$? – StubbornAtom Dec 3 '19 at 14:58