The link between large p-values and low power? In Wasserman's book "All of Statistics", page 157, there's the following sentence:
«A large p-value is not strong evidence in favor of $H_0$. A large p-value can occur for two reasons: 


*

*$H_0$ is true, or

*$H_0$ is false and the test has low power.»


Why when we have a low power test, do we have a large p-value?
Any help would be appreciated. 
 A: First thing, power of the test $\delta(X)$ is a function $1 - \beta(\theta)$ where
$$\beta(\theta) = P_{\theta}(\delta(X) = 0) \text{ - probability of accepting $H_0$}$$
So if we have low power and $H_0$ is false, i.e. $1 - \beta(\theta)$ is small for $\theta_1$, then probability of accepting $H_0$ even if it is false is equal to $\beta(\theta_1)$ and is high. If our test has a form: $1[T(X) \ge c_{1 - \alpha}]$ for $c_{1 - \alpha}$ being $1 - \alpha$ quantile of $T(X)$, then
$$P_{\theta_1}(T(X) \ge c_{1 - \alpha}) \text{ is close to 0}$$
Now we know that power of every "decent" test is not lower than its significance level (test is unbiased), thus
$$P_{\theta_1}(T(X) \ge c_{1 - \alpha}) \ge P_{\theta_0}(T(X) \ge c_{1 - \alpha}) \text{ is close to 0}$$
And equivalently, p-value of the test $P \ge \alpha$ almost surely. So, does it imply $\textbf{high}$ $P$? We only know it's larger than $\alpha$ but it doesn't have to be close to $1$.
As to better see why large p-value $P$ means nothing, even assuming $H_0$ to be true, it is known that
$$P_{\theta_0}(P \le q) = q$$
$$$$ So under $H_0$ p-value is distributed $\textbf{uniformly}$ on $[0,1]$. So all that really matters is if equality $P < \alpha$ is true or not, and to the exact value of $P$.
