Dapatkah Neyman-Pearson lemma berlaku untuk kasus ketika null sederhana dan alternatif tidak termasuk dalam keluarga distribusi yang sama?

1. Dapatkah lemma Neyman-Pearson berlaku untuk kasus ketika nol sederhana dan alternatif sederhana tidak termasuk dalam keluarga distribusi yang sama? Dari buktinya, saya tidak mengerti mengapa itu tidak bisa.

   Misalnya, ketika nol sederhana adalah distribusi normal dan alternatif sederhana adalah distribusi eksponensial.

2. Apakah uji rasio kemungkinan merupakan cara yang baik untuk menguji komposit nol terhadap alternatif komposit ketika keduanya milik keluarga distribusi yang berbeda?

Terima kasih dan salam!

Ya Neyman Pearson Lemma dapat berlaku untuk kasus ini ketika null sederhana dan alternatif sederhana tidak termasuk dalam keluarga distribusi yang sama.

Mari kita ingin membangun tes Paling Kuat (MP) dari terhadap H 1 : X ∼ Exp ( 1 $H_0:X\sim N(0,1)$$H_1 : X\sim \text{Exp}(1)$ dari ukurannya.

Untuk tertentu , fungsi kritis kami oleh Neyman Pearson lemma adalah$k$

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

adalah tes MP dari terhadap H $H_0$$H_1$ dari ukurannya.

Di sini

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

Perhatikan bahwa Sekarang jika Anda menggambar gambarr(x)[Saya tidak tahu bagaimana membuat gambar sebagai jawaban], dari grafik akan jelas bahwar(x)>

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

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

• 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.

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.