# Can Neyman-Pearson lemma apply to the case when simple null and alternative don’t belong to the same family of distributions?

1. Can the Neyman-Pearson lemma apply to the case when a simple null and
a simple alternative don’t belong to the same family of distributions?
From its proof, I don’t see why it can’t.

For example, when the simple null is a normal distribution and the
simple alternative is a exponential distribution.

2. Is the likelihood ratio test a good way to test a composite null against a
composite alternative when both belong to different families of
distributions?

Thanks and regards!

Yes Neyman Pearson Lemma can apply to the case when simple null and simple alternative don’t belong to the same family of distributions.

Let we want to construct a Most Powerful(MP) test of $H_0:X\sim N(0,1)$ against $H_1 : X\sim \text{Exp}(1)$ of its size.

For a particular $k$, our critical function by Neyman Pearson lemma is

is a MP test of $H_0$ against $H_1$ of its size.

Here

Note that
Now if you draw the picture of $r(x)$ [I don’t know how to construct a Picture in answer ], from graph it will be clear that $r(x)>k \implies x>c$.

So, for a particualr $c$

is a MP test of $H_o$ against $H_1$ of its size.

You can test

1. $H_0:X\sim N(0,\dfrac{1}{2})$ against $H_1:X\sim \text{Cauchy}(0,1)$
2. $H_0:X\sim N(0,1)$ against $H_1:X\sim \text{Cauchy}(0,1)$
3. $H_0:X\sim N(0,1)$ against $H_1:X\sim \text{Double Exponential}(0,1)$

By Neyman Pearson lemma.

Normally the likelihood ration test(LRT) is not a good way for composite null and composite alternative which belong to different family of distributions.The LRT is specially useful when $\mathbb{\theta}$ is a multi-parameter and we wish to test hypothesis concerning one of the parameters.

That’s all from me.