# Uniformly Most Powerful test for normal distribution

Let ${Y_1,...,Y_n}$ be independent random variables and $Y_i$~$N(\beta x_i, 1)$ where $x_1,...,x_n$ are fixed known constants, and $\beta$ is an unknown parameter.

I'm trying to find a uniformly most powerful level $\alpha$ test for

$$H_0: \beta=0 \quad \text{vs} \quad H_1:\beta>0$$

First I found the Likelihood Ratio statistic

\begin{align}\frac{L(0)}{L(\beta_1)}&=\frac{e^{-\frac{1}{2}\sum^n_{i=1}(y_i-0x_i)^2}}{e^{-\frac{1}{2}\sum^n_{i=1}(y_i-\beta_1x_i)^2}} \\[15pt] & = \frac{e^{-\frac{1}{2}\sum^n y_i^2}}{e^{-\frac{1}{2}(y_i^2-2y_i\beta_1x_i+\beta_1^2x_i^2)}} \\[15pt] & =e^{\beta_1\sum^n y_ix_i-\frac{1}{2}\beta_1^2\sum^nx_i^2}\end{align}

Then the rejection region of the LRT is $$e^{\beta_1\sum^n y_ix_i-\frac{1}{2}\beta_1^2\sum^nx_i^2} \le k \\$$

Now moving the constants to the right side $$\\[15pt]\Rightarrow\beta_1\sum^n y_ix_i-\frac{1}{2}\beta_1^2\sum^nx_i^2 \;\le\; \text{ln}(k)$$

$$\\[15pt]\Rightarrow\sum^ny_ix_i \;\le\; \frac{\text{ln}(k)+1/2\sum^n\beta_1^2x_i^2}{\beta_1}$$

Here is where I get confused because I'm not sure what to leave on the left side. From what I have read I'm supposed to move constants to the right. I'm thinking that only $y_i$'s should remain on the left. Since $x_i$'s are constants shouldn't I move it?

The steps that follow to find the rejection region by using the distribution of the left of the inequality, then equating to a quantile, puzzle me because of the $x_i$

I'm not even sure I'm going about this correctly or if there are more straight forward methods.

First of all, you lose a sign: \begin{align}\frac{L(0)}{L(\beta_1)}&=\frac{e^{-\frac{1}{2}\sum^n y_i^2}}{e^{-\frac{1}{2}(y_i^2-2y_i\beta_1x_i+\beta_1^2x_i^2)}} \\[15pt] & =e^{-\beta_1\sum^n y_ix_i+\frac{1}{2}\beta_1^2\sum^nx_i^2}\end{align} The rejection region looks as $$-\beta_1\sum_{i=1}^n y_ix_i+\frac{1}{2}\beta_1^2\sum_{i=1}^nx_i^2 \leq k.$$ Since only $y_i$ are random variables here, and $k$ should be found so that test has level $\alpha$, simplify rejection region: $$\sum_{i=1}^n y_ix_i \geq k_0.$$

Let us find $k_0$ from type 1 error $\alpha_1=\alpha$. We need to know the distribution of l.h.s. under $H_0$: $$\sum_{i=1}^n y_ix_i \sim N\bigl(0, \sum_{i=1}^n x_i^2\bigr).$$ Then $$\alpha_1=\mathbb P_{H_0}\left(\sum_{i=1}^n y_ix_i \geq k_0\right)=P_{H_0}\left(\frac{\sum_{i=1}^n y_ix_i}{\sqrt{\sum_{i=1}^n x_i^2}} \geq \frac{k_0}{\sqrt{\sum_{i=1}^n x_i^2}}\right)=1-\Phi\left(\frac{k_0}{\sqrt{\sum_{i=1}^n x_i^2}}\right)=\alpha$$ if $$\frac{k_0}{\sqrt{\sum_{i=1}^n x_i^2}} = z_{1-\alpha},\ \quad k_0 = z_{1-\alpha}\cdot\sqrt{\sum_{i=1}^n x_i^2},$$ where $z_{1-\alpha}$ is the $(1-\alpha)^{\text{th}}$ quantile of standard normal distribution.

Finally, we've got rejection region of level $\alpha$ test for testing $H_0$ under the simple alternative $H_1: \beta=\beta_1>0$: $$\sum_{i=1}^n y_ix_i \geq z_{1-\alpha}\cdot\sqrt{\sum_{i=1}^n x_i^2}.$$ Since this test

1) is the most powerful test for any $\beta_1>0$,

2) does not depend on $\beta_1>0$ at all

is is the uniformly most powerful test for testing $H_0$ under the composite alternative $H_1: \beta>0$.

• This was very helpful, thank you!
– Mike
Commented May 18, 2017 at 17:39
• After you ask a question here, if you get an acceptable answer, you can "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?.
– NCh
Commented May 21, 2017 at 13:02