Power function in hypothesis testing

Question

Example 1:

$X_1,...,X_n$ is a random sample from $\text{Poisson}(\lambda)$

$T=\sum X_i$

$T\sim \text{Poisson}(n\lambda)$

Consider the test: $H_0: \lambda \le \lambda_0$ vs $H_1: \lambda > \lambda_0$

The rejection region is of the form $T>c$

Let $\lambda_0=1$ and use CLT to choose $n$ so that the power function $\beta(\lambda)$ satisfies $\beta(1)=0.01$ , $\beta(2)=0.05$. Specify also the "rejection limit" $c$ which you find.

Note: if $Z\sim N(0,1)$ then $P(Z\ge 1.645)=0.05$ and $P(Z\ge 2.326)=0.01$

Solution:

We have $ET=\text{Var}(T)=n\lambda$ and it follows from CLT that if $n$ is large

$\frac{T-n\lambda}{\sqrt(n\lambda)}\sim N(0,1)$

we have

$\beta(1)=P(T>c|\lambda=1)=P(\frac{T-n}{\sqrt{n}}>\frac{c-n}{\sqrt{n}}|\lambda=1)=P(Z>\frac{c-n}{\sqrt{n}})=0.01$

$\beta(2)=P(T>c|\lambda=2)=P(\frac{T-2n}{\sqrt{2n}}>\frac{c-2n}{\sqrt(2n)}|\lambda=2)=P(Z>\frac{c-2n}{\sqrt{2n}})=0.95$

which gives us the equations

$c-n=2.326\sqrt{n}$

$c-2n=-1.645\sqrt{2n}$

which has the solutions $n=21.64$, $c=32.47$ or n=22 and c=33

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Example 2:

Let $X_1,...,X_n$ be a random sample from $N(\theta,\sigma ^2)$

$H_0: \theta \le \theta_0 $ vs $H_1: \theta > \theta_0$

reject $H_0$ if $\frac{\bar X-\theta_0}{\sigma /\sqrt {n}}>c$, where $c$ is any positive number

The power function of the test is:

$\beta(\theta)=P_\theta(\frac{\bar X-\theta_0}{\sigma /\sqrt n}>c) = P_\theta(\frac{\bar X-\theta_0}{\sigma /\sqrt n}>c +\frac{\theta_0-\theta}{\sigma /\sqrt n})=P(Z>c+\frac{\theta_0-\theta}{\sigma /\sqrt n})$

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Question: Why does the power function become like that? when they are finding $\beta(1), \beta(2)$ etc? The "logic" in example 1 doesn't really seem to match up with the "logic" in example 2? Could someone explain this?

Additionally I am not really understanding the equality in the last line in example 2: basically how: $P_\theta(\frac{\bar X-\theta_0}{\sigma /\sqrt n}>c) = P_\theta(\frac{\bar X-\theta_0}{\sigma /\sqrt n}>c +\frac{\theta_0-\theta}{\sigma /\sqrt n})$.

Answer 1

I will start from the last question and work backwards. I think there might be a typo in the book or in your transcription:

\begin{align} P_\theta\left(\frac{\bar X-\theta_0}{\sigma /\sqrt n}>c\right) & = P_\theta\left(\bar X > \theta_0 + c \, \sigma /\sqrt n\right) \\ & = P_\theta\left(\bar X - \theta > \theta_0 - \theta + c \, \sigma /\sqrt n\right) \\ & = P_\theta\left(\frac{\bar X-\theta}{\sigma /\sqrt n} > c +\frac{\theta_0-\theta}{\sigma /\sqrt n}\right) \\ & = P_\theta\left(Z > c +\frac{\theta_0-\theta}{\sigma /\sqrt n}\right) \\ & = 1-\Phi\left(c +\frac{\theta_0-\theta}{\sigma /\sqrt n}\right) \end{align}

The point is that, you are dealing with a general expression for the probability of rejecting the null hypothesis at any $\theta$ in the parameter space and from the final expression you can observe that this expression, as a function of $\theta$, is an increasing function. When $\theta=\theta_0$ then, we have the $\sup$ of this function over the null parameter space, $\sup = 1-\Phi(c)$. For a specified significance level $\alpha$ then, we will take $c = \Phi^{-1}(1-\alpha)$ guaranteeing the worst Type-I error to be no more than $\alpha$.

Moving on to the first example, note that you are dealing with $\sum{X_i}$ versus $\bar X$ in the second example. Also, Poisson has mean = variance.

You may change the 1st example to be similar to the 2nd by saying: "Reject null if:"

\begin{align} T & > c \\ \sum{X_i} & > c \\ \bar X & > \frac{c}{n} \\ \frac{\bar X - \lambda_0}{\sqrt{\lambda_0}/ \sqrt n} & > \frac{\frac{c}{n} - \lambda_0}{\sqrt{\lambda_0}/ \sqrt n} \end{align}

But this point, we should realize that writing an expression like example 2 is not easy because $\lambda_0$ appears in both numerator and denominator.

So it is easier to deal with $\sum{X_i} > c$. So we get

\begin{align} P_\lambda\left(\sum{X_i} > c \right) & = P_\lambda\left( \frac{\sum{X_i} - n \lambda }{\sqrt{ n \lambda}} > \frac{c - n \lambda }{\sqrt{ n \lambda}}\right) \\ & \overset{\mathrm{CLT}}{\approx} P_\lambda\left(Z > \frac{c - n \lambda }{\sqrt{ n \lambda}}\right) \end{align}

In addition, problem 1 gives the values of the power function at two specific parameter values and asks you to solve for two unknowns $c$ and $n$ given those. One could ask a similar question in example 2 also, but there the problem will need to provide $\sigma$ in addition.