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


*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?
 A: Q2. The likelihood ratio's a sensible enough test statistic but (a) the Neyman-Pearson Lemma doesn't apply to composite hypotheses, so the LRT won't necessarily be most powerful; & (b) Wilks' Theorem only applies to nested hypotheses, so unless one family is a special case of the other (e.g. exponential/Weibull, Poisson/negative binomial) you don't know the distribution of the likelihood ratio under the null, even asymptotically.
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*You're exactly right.  The general picture is: we want a test statistic that gives us maximal power at a given significance level $\alpha$.  In other words, a way to compute a value $\phi$ so that the points part of parameter space for which $\phi$ exceeds its $\alpha^\mathrm{th}$ quantile under $H_0$ have the least possible weight under $H_1$.  The Neyman-Pearson lemma demonstrates that that statistic is the likelihood ratio.


*Neyman & Pearson's original paper also discusses composite hypotheses. In some cases the answer is straightforward -- if there is a choice of particular distributions in each family whose likelihood ratio is conservative when applied the the whole family.  This is what often happens, for instance, for nested hypotheses.  It's easy for this not to happen, though; this paper by Cox discusses what to do further.  I think a more modern approach here would be to approach it in a Bayesian way, by putting priors over the two families.
A: 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 
$$\phi(x) =\begin{cases} 1,&\dfrac{f_1(x)}{f_0(x)}>k \\0, &\text{Otherwise} \end{cases}$$ 
is a MP test of $H_0$ against  $H_1$ of its size. 
Here $$r(x)=\dfrac{f_1(x)}{f_0(x)}=\dfrac{e^{\displaystyle -x}}{\frac{1}{\sqrt{2 \pi}}e^{\displaystyle -x^2/2}}=\sqrt{2 \pi}\,\,e^{\displaystyle \left(\frac{x^2} 2-x\right)}$$
Note that  $$r'(x) =\sqrt{2 \pi}\,\,e^{\displaystyle \left(\frac{x^2} 2-x\right)}(x-1)\\ \begin{cases}<0 ,& x<1\\>0 ,& x>1 \end{cases}$$
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 $c$
$$\phi(x) =\begin{cases} 1,&x>c \\0, &\text{Otherwise} \end{cases}$$ 
is a MP test of $H_o$ against $H_1$ of its size.
You can  test 


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

*$H_0:X\sim N(0,1)$ against $H_1:X\sim \text{Cauchy}(0,1)$

*$H_0:X\sim N(0,1)$ 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.
