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 $P(Z\ge 2.326)=0.01$


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



which gives us the equations



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

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:

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

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

  • $\begingroup$ You'll need to look up (in the unidentified text) and supply definitions of the terms you're asking about. $\endgroup$
    – Glen_b
    May 17, 2017 at 0:04
  • $\begingroup$ It's not clear to me. As one example, why is the symbol ($\beta$) which usually represents the type II error rate being referred to as a power function, when normally $\beta$ is the complement of power? What's its mathematical definition? $\endgroup$
    – Glen_b
    May 17, 2017 at 1:04
  • $\begingroup$ another issue is that you didn't clearly identify what the distinction between the two examples is that bothers you. You juxtaposed them but didn't make it clear what it is that you find troublesome, making it hard to explain the problem. I will try cleaning up your mathematics a bit, maybe that will help a little $\endgroup$
    – Glen_b
    May 17, 2017 at 1:07
  • $\begingroup$ Where do these come from? Are they in a book? $\endgroup$
    – Glen_b
    May 17, 2017 at 1:15
  • 1
    $\begingroup$ @Glen_b: Unfortunately a number of math stat textbooks (e.g., DeGroot and Schervish) use $\beta(\theta)$ to represent the power function, i.e., $\beta(\theta) = P_{\theta}(\text{Reject }H_0)$ for any $\theta$ in $\Omega$ the parameter space. Then the probability of type-II error (which is called $\beta$ commonly by practitioners as you say) for some $\theta_1 \in \Omega_1$ the parameter space defined by the alternative hypothesis is $1-\beta(\theta_1)$, while what is called $1-\beta=$ Power by practitioners is $\beta(\theta_1)$. An unfortunate choice of notation, indeed! $\endgroup$ May 18, 2017 at 3:25

1 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.


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