Neyman-Pearsonの補題は、単純なヌルと代替が同じ分布ファミリーに属さない場合に適用できますか？

1. Neyman-Pearson補題は、単純なヌルと単純な代替が同じ分布ファミリーに属さない場合に適用できますか？その証拠から、なぜできないのかわかりません。

   たとえば、単純なヌルが正規分布であり、単純な代替が指数分布である場合。

2. 尤度比テストは、両方が異なる分布のファミリーに属している場合、複合代替に対して複合ヌルをテストする良い方法ですか？

よろしくお願いします！

はい、Neyman Pearson Lemmaは、単純なヌルと単純な代替が同じ分布ファミリーに属さない場合に適用できます。

私たちは最も強力な（MP）の試験構築したいましょうに対するH 1：X 〜経験（1 ）そのサイズのを。$H_0:X\sim N(0,1)$$H_1 : X\sim \text{Exp}(1)$

特定のについて、Neyman Pearson補題による臨界関数は$k$

$$\phi(x) =\begin{cases} 1,&\dfrac{f_1(x)}{f_0(x)}>k \\0, &\text{Otherwise} \end{cases}$$

MPの試験であるに対する H 1の大きさの。$H_0$$H_1$

ここで、

$$r(x)=\dfrac{f_1(x)}{f_0(x)}=\dfrac{e^{\displaystyle -x}}{\frac{1}{\sqrt{2 \pi}}e^{\displaystyle -x^2/2}}=\sqrt{2 \pi}\,\,e^{\displaystyle \left(\frac{x^2} 2-x\right)}$$

なお、

$$r'(x) =\sqrt{2 \pi}\,\,e^{\displaystyle \left(\frac{x^2} 2-x\right)}(x-1)\\ \begin{cases}<0 ,& x<1\\>0 ,& x>1 \end{cases}$$ 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 $r(x)>k \implies x>c$.

$c$

$$\phi(x) =\begin{cases} 1,&x>c \\0, &\text{Otherwise} \end{cases}$$ $H_o$$H_1$

テストできます

• 1. $H_0:X\sim N(0,\dfrac{1}{2})$$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.

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.

1. 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.

2. 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.