# Hypothesis Testing for different distribution

Assume that $$X$$ is a random variable from a distribution with density function $$f$$. Find the most powerful test of size-$$\alpha$$ to test $$H_0:X\sim G(1,1)$$ $$H_1: X\sim N(1,1)$$

I know that I need to use Neyman Pearson Lemma usually to solve this kind of question. But normally I am given the parameter (like $$\theta$$) for this type of questions. But now I am not really sure how to proceed.

I would appreciate any help or hints.

• What is $G$?$\text{}$ Commented Jan 11, 2020 at 12:04
• $G$ is gamma dsitribution, $N$ is normal distribution Commented Jan 11, 2020 at 12:10
• The method is same: You find the likelihood ratio $f_{H_1}/f_{H_0}$ and reject $H_0$ for large values of this ratio. Commented Jan 11, 2020 at 12:41
• that's mean I just substitute 1 into the parameter and evaluate them? Commented Jan 11, 2020 at 13:20

You can think of the situation like this:

$$X\sim \theta \,\Gamma(1,1)+(1-\theta)N(1,1)\,,\quad\theta\in\{0,1\}$$

You are to test $$H_0:\theta=1$$ against $$H_1:\theta=0$$, i.e. a simple null versus a simple alternative.

But to actually solve the problem, there is no need to introduce any parameter $$\theta$$.

Hint:

Follow the standard procedure. Let $$f$$ be the density of $$X$$. Find the likelihood ratio $$f_{H_1}/f_{H_0}$$, where $$f_{H_j}$$ is the pdf of $$X$$ under $$H_j$$, $$j=0,1$$. By Neyman-Pearson lemma, a most powerful test rejects $$H_0$$ whenever the ratio $$f_{H_1}/f_{H_0}$$ is large. Simplify the rejection region from there and hence find the exact test.

• Ya, this is the standard procedure. After I evaluated their density function, I just substitute $\alpha=1$ ,$\lambda=1$, $\mu = 1$, and $\sigma^2=1$ into the equation right? Commented Jan 11, 2020 at 14:18
• If they are the parameters of your distributions, yes. Commented Jan 11, 2020 at 14:22