Statistical hypothesis testing - how to demonstrate that power $>$ size Let $(X_1,X_2,\ldots,X_n)$ be a random sample from $N(\mu,\sigma^2)$, where $\sigma$ is known. Find the power function of the test of significance of $H_0:\mu=\mu_0$ against $H_1:\mu\neq\mu_0$ . Also, show that power${}>{}$size. My first step is to define the critical region as $W=\{(x_1,x_2,\ldots,x_n):|\sqrt{n}(\bar x-\mu_0)/\sigma|>\tau _{\alpha/2}\}$ where $P[Z>\tau_\alpha]=\alpha$, $Z\sim N(0,1)$, and $\alpha$ is the level of significance.
The power function of the test of significance is then, 
$\beta(\mu)=P[(X_1,\ldots,X_n)\in W|\mu\in\Omega_1]=P[|\sqrt{n}(\bar x-\mu_0)/\sigma|>\tau _{\alpha/2}|\mu\neq\mu_0]$
and the size of the test of significance is,
$\sup\limits_{\mu \in\Omega_0}P[ (X_1,\ldots,X_n)\in W|\mu\in\Omega_0]=P[|\sqrt{n}(\bar x-\mu_0)/\sigma|>\tau _{\alpha/2}|\mu=\mu_0]$.
How do I proceed from here?
 A: Your notation $P[(X_1,\ldots,X_n)\in W|\mu\in\Omega_1]$ has no sense. It is true that one commonly says that the power is the probability to reject under $H_1$ (here $H_1=\{\mu \in \Omega_1\}$ but unfortunately this has no sense because there are several possible probabilities under $H_1$.
The power function is $\mu \mapsto P[(X_1,\ldots,X_n)\in W|\mu]$ when we consider that $(X_1, \ldots, X_n)$ is a i.i.d.\ sample of ${\cal N}(\mu, \sigma^2)$ under the probability $P[\cdot|\mu]$. Rigorously we may consider that the power is not defined for $\mu=\mu_0$ but this has no importance.
Finally to answer your question you have to show that the distribution of the test statistic $\left|\sqrt{n}\frac{\bar X - \mu_0}{\sigma}\right|$ under $P(\cdot |\mu)$ is stochastically increasing in $\mu$. This tantamounts to say that the distribution of $\left|\bar X - \mu_0\right|$ under $P(\cdot |\mu)$ is stochastically increasing in $\mu$ when $\mu > \mu_0$ and stochastically decreasing in $\mu$ when $\mu < \mu_0$.
A: Note that  $\beta(\mu)=P[(X_1,\ldots,X_n)\in W|\mu\in\Omega_1]=P[|\sqrt{n}(\bar X-\mu_0)/\sigma|>\tau _{\alpha/2}]$$=P[\sqrt{n}(\bar X-\mu_0)/\sigma>\tau _{\alpha/2}$ or $<-\tau_{\alpha/2}$]$=P[\sqrt{n}(\bar X-\mu)/\sigma+\sqrt{n}(\mu-\mu_0)/\sigma>\tau _{\alpha/2}$ or $ < -\tau _{\alpha/2}]$$$=P[\sqrt{n}(\bar X-\mu)/\sigma>\tau _{\alpha/2}+\sqrt{n}(\mu_0-\mu)/\sigma]+P[\sqrt{n}(\bar X-\mu)/\sigma>-\tau _{\alpha/2}+\sqrt{n}(\mu_0-\mu)/\sigma]$$$$=1-\Phi(\tau _{\alpha/2}+\sqrt{n}(\mu_0-\mu)/\sigma)+\Phi(-\tau _{\alpha/2}+\sqrt{n}(\mu_0-\mu)/\sigma)$$Now differentiate $\beta(\mu)$ w.r.t $\mu$ we get $$\beta'(\mu)=[\phi(\tau _{\alpha/2}+\sqrt{n}(\mu_0-\mu)/\sigma)-\phi(-\tau _{\alpha/2}+\sqrt{n}(\mu_0-\mu)/\sigma)]\frac{\sqrt{n}}{\sigma}$$ Now equating $\beta'(\mu)=0$ we get $$\phi(\tau _{\alpha/2}+\sqrt{n}(\mu_0-\mu)/\sigma)=\phi(\tau _{\alpha/2}+\sqrt{n}(\mu-\mu_0)/\sigma)$$ [Since $\phi (-x)=\phi(x)$]$$=>\mu=\mu_0$$ Note that if $\mu<\mu_0$ then $\beta'(\mu)<0 $ and if $\mu>\mu_0$ then $\beta'(\mu)>0 $ So from the first order derivative test we get  $\beta(\mu)$ is minimum at $\mu=\mu_0$. So,hence it is proved that power > size
