Neyman-Pearson Lemma

Best Critical Region: Let $C$ be a critical region for a hypothesis test with significance level, where the hypotheses are $H_0:\theta ={\theta }_0$ and $H_1:\theta ={\theta }_1$. Then we say that $C$ is a best critical region of size $\alpha$ if whenever $D$ is another critical region with size $\alpha$, then $P(C\mid \theta ={\theta }_1)\ge P(D\mid \theta ={\theta }_1)$.

What this means is that if the alternative hypothesis is true, then the probability that we reject the null hypothesis is greatest if we use the critical region $C$. Besides, this best critical regions is the most powerful among all the regions with size $\alpha$, and thus the UMPT.

Neyman-Pearson Lemma: Let $X_1,\dots ,X_n$ be a random sample sample from a distribution with PDF $f(x;\theta )$, and let the likelihood function $\mathcal{L}(\theta )=\prod f(x;\theta )$. If there exists a positive constant $k$ and a region $C$ such that

1. $P[(X_1,\dots ,X_n)\in C\mid \theta ={\theta }_0]=\alpha$
2. $\frac{\mathcal{L}({\theta }_0)}{\mathcal{L}({\theta }_1)}\le k$ for $(x_1,\dots ,x_n)\in C$ and $\frac{\mathcal{L}({\theta }_0)}{\mathcal{L}({\theta }_1)}\ge k$ for $(x_1,\dots ,x_n)\in C^{\prime }$ Then $C$ is a best critical region of size $\alpha$ for testing $H_0:\theta ={\theta }_0$ against $H_1:\theta ={\theta }_1$

Example: Let $X_1,\dots ,X_n$ be a random sample from $\mathcal{N}(\mu ,64)$.

1. Show $C=\left\{(x_1,\dots ,x_n)\mid \bar{x}\le c\right\}$ is a best critical region for testing $H_0:\mu =80$ against $H_1:\mu =76$.
2. Find $n$ and $c$ so that $\alpha \approx 0.05$ and $\beta \approx 0.05$.

First we calculate

$$\begin{array}{rl}\frac{\mathcal{L}(80)}{\mathcal{L}(76)}& =\frac{\prod \frac{1}{8\sqrt{2\pi }}\exp \left(-(x_i-80{)}^2/128\right)}{\prod \frac{1}{8\sqrt{2\pi }}\exp \left(-(x_i-76{)}^2/128\right)}\\ & =\exp \left(-\frac{1}{128}\sum \left({\left(x_i-80\right)}^2-{\left(x_i-76\right)}^2\right)\right)\\ & =\exp \left(-\frac{1}{128}\left(-8\sum x_i+624n\right)\right)\end{array}$$

We need this ratio to be less than a positive constant $k$

$$\exp \left(-\frac{1}{128}\left(-8\sum x_i+624n\right)\right)\le k⟹\bar{x}\le \frac{16\ln k+78n}{n}$$

for $(x_1,\dots ,x_n)\mid \bar{x}\le c$, so we can just let $c=\frac{16\ln k+78n}{n}$.

We want $P(\bar{x}\le c\mid \mu =80)=0.05$ and $P(\bar{x}>c\mid \mu >c)=0.05$, that is

$$\frac{c-80}{8/\sqrt{n}}=-1.645,\frac{c-76}{8/\sqrt{n}}=1.645$$

where $\frac{8}{\sqrt{n}}=\frac{\sigma }{\sqrt{n}}=\mathrm{SE}$ is the standard error, and $\Phi (1.645)=0.95,\Phi (-1.645)=0.05$. Then we could solve for $c=78;n\approx 43$.