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Let $X_i\sim N(\mu,\sigma^2)$, for known $\sigma^2$. We want to test $H_0: \mu=\mu_0$ vs $H_1: \mu>\mu_0$.

By Karl-Rubin theorem, we know that the UMP test is given by the rejection region $W$ defined by $T(X_1,...,X_n)>t_0$, when the ratio $\frac{f(x_1,...,x_n|\mu)}{f(x_1,...,x_n|\mu_0)}$ is a non-decreasing function of $T(X_1,...,X_n)$, for a given size $\alpha$.

In our case $T(X_1,...,X_n)=\bar{X}$. So, $\bar{X}>t_0\Leftrightarrow \frac{f(x_1,...,x_n|\mu)}{f(x_1,...,x_n|\mu_0)}>C \Leftrightarrow(x_1,...,x_n)\in W$

The power function is the $\beta(\mu)=P((x_1,...,x_n)\in W|\mu)=P(\bar{X}>t_0|\mu)=1-\Phi(\frac{t_0-\mu}{\sigma/\sqrt{n}})$.

We fix $t_0$ by solving $\alpha=P((x_1,...,x_n)\in W|\mu=\mu_0)$.

Now my doubt is about the interplay between $t_0, C, W$ and $\mu$. How can we keep $t_0$ fixed, and vary $\mu$ for a plot of the power function? Even if it's possible, aren't we changing $C$ and consequently $W$, by varying $\mu$?

I would say that if we keep $t_0$ fixed, and vary $\mu$, then we'll varying $C$ and $W$, and that's why by increasing $\mu$ in our case, we'll be increasing the power, i.e. augmenting the rejection region $W$. Am I correct?

Any help would be appreciated

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The quantities $t_0$, $C$ and $W$ are all deterministically linked with $\mu_0$. They all demarcate the same rejection region that has $\alpha$ probability if $\mu = \mu_0$. That's a big if. It's all the same data region, just stated in different ways.

Now, given these quantities, what's the probability of falling into the rejection region if the mean parameter equals a general $\mu$? Well it's not going to be $\alpha$ anymore (unless $\mu$ is exactly $\mu_0$ but that's probably not gonna happen). That's what the power function gives you.

You might be confused by this $$ \frac{f(x_1,...,x_n|\mu)}{f(x_1,...,x_n|\mu_0)}>C $$ because it has a $\mu$ written in the numerator. But you're just writing it in a weird way or equivocating on what $\mu$ is. This quantity should be a statistic, and not depend on unknown parameters. Otherwise how could you do the test? Right?


Here's an example:

Say you have normal data with known $\sigma^2$, and you are testing $H_0: \mu = \mu_0$ versus $H_0:\mu > \mu_0$. Pick any $\mu_a > \mu_0$. Your likelihood ratio is $$ \exp\left[-\frac{1}{2\sigma^2}\left(\sum(x_i - \mu_a)^2 - \sum(x_i - \mu_0)^2)\right) \right] = \exp\left[-\frac{1}{2\sigma^2}(n\mu_a^2 - n\mu_0^2) \right]\exp\left[-\frac{1}{2\sigma^2}\sum x_i(\mu_0- \mu_a) \right]. $$ You reject when that's bigger than some $C$. It would be equivalent to say to reject the test when $$ \exp\left[-\frac{1}{2\sigma^2}\sum x_i(\mu_0- \mu_a) \right] $$ is bigger than some different number, call it $C'$ (we divided both sides of an inequality by a positive number, right?). Taking log of both sides, and dividing by another positive number, we see this is the same as rejecting when $\sum x_i$ is ``large", greater than some number $C''.$ Go a bit further: it's the same as rejecting when $\bar{X} > C''' = t_0$. These are all precisely the same event. They are all logically equivalent. We just did algebra kept rewriting the same thing.

We keep simplifying because it is easy to do the next step if the statistic has a nice distribution. We pick this number according to $\mu_0$ so that the probability of it happening, if $\mu = \mu_0$ is true, will have the right type 1 error. Notice that it will be the same rejection region no matter what $\mu_a$ is! For this particular example, $t_0$ is the $(1-\alpha)100$th percentile of a $\text{Normal}(\mu_0, \sigma^2/n)$ distribution.

But then what's the probability of this happening once we pick $t_0$? What's $$ P_{\mu}(\bar{X} \ge t_0)? $$ Well $\bar{X} \sim \text{Normal}(\mu, \sigma^2/2)$, and now we don't assume the null is true. This depends on what the true $\mu$ is. So the probability changes with $\mu$. However the event does not.

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  • $\begingroup$ Taylor, thanks for you answer. I still have my doubts though. Let's work backwards. Let's fix $t_0$. For us to go from $C'''$ to $C$, at some point we do an operation dependent on your $\mu_a$, namely from $C''$ to $C'$. If the sample is fixed, then changing $\mu_a$, ultimately will also lead to a different $C$, since all the other operations are still the same. If $C$ changes, doesn't also $W$? $\endgroup$ – An old man in the sea. Jan 27 '17 at 9:05
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    $\begingroup$ Actually yeah that's a good point. $C$ depends on the alternative parameter. $W$ can't though--it won't be a hypothesis test then. The KR theorem is basically saying it doesn't matter what it is though, because the inequality is always pointing the same way after all the steps. And the cut off region is always the same. So you can just use Neyman Pearson for any $\mu_a$ $\endgroup$ – Taylor Jan 27 '17 at 14:50
  • $\begingroup$ taylor, could you please check if my answer is correct? $\endgroup$ – An old man in the sea. Jan 27 '17 at 16:46
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I've been thinking about this. Here's what I think is happening.

$P(\bar{X}>k|\mu=\mu_0)=\alpha$ defines a specific rejection region $W$ and $k$.

And when, for the power function, we vary $\mu_1$, we're maintaining the same rejection region $W$, precisely because for each $\mu_1$, we're changing $C$ such that $\bar{X}>k\Leftrightarrow (X_1,...,X_n)\in W$ for exactly the same $k$ as before.

We maintain $W$ and $k$, by varying $C$ according to $\mu_1$

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    $\begingroup$ Yes I think this is correct $\endgroup$ – Taylor Jan 27 '17 at 18:19

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