Suppose $X_{i}...X_{n}$ are i.i.d Bernoulli with $P_{\theta}(X_{i} = 1) = \theta = 1 - P_{\theta}(X_{i} = 0)$.
n is large. Consider testing $H_{0} = 0.5$ versus $H_{1} : \theta > 0.5.$
Let $\hat \theta_{n} = \sum \frac{X_{i}}{n}$ be the MLE of $\theta$. Consider the large sample test which rejects when $2\sqrt{n}(\hat \theta_{n} - 0.5) > z_{1-\alpha}$, where $z_{1-\alpha}$ is the 1 - $\alpha$ quantile of the standard normal distribution.
(i) An approximate p-value can be computed as the area under the normal curve to the right of $2\sqrt{n}(\hat \theta_{n} - 0.5)$. We can define the p-value $\hat p_{n}$ by $1 - \Phi[2\sqrt{n}(\hat \theta_{n} - 0.5)]$. Where $\Phi$ denotes that standard normal distribution. If $H_{0}$ is true, what is the limiting distribution of $\hat p_{n}$?
Here is what I believe I know if $H_{0}$ is true:
$2\sqrt{n}(\hat \theta_{n} - 0.5)$ $\xrightarrow{d}$ N(0,1) (MLE asymptotic normality)
$\therefore$ $\hat p_{n}$ $\xrightarrow{d}$ U[0,1]
If $H_{a}$ is true, here is what I have gotten (Where $\theta_{1}$ is the $H_{a}$ parameter and $\theta_{0}$ is the $H_{0}$ parameter):
$2\sqrt{n}(\hat \theta_{n} - (0.5 + \theta_{1} - \theta_{0})$ $\xrightarrow{d}$ N(0,1)
$\therefore$ $2\sqrt{n}(\hat \theta_{n} - 0.5)$ $\xrightarrow{d}$ N($2\sqrt{n}(\theta_{1} - \theta_{0}),1)$
If Z = $2\sqrt{n}(\hat \theta_{n} - (0.5 + \theta_{1} - \theta_{0}))$, then:
P(Z $\geq z_{1-\alpha} - 2 \sqrt{n}(\theta_{1} - \theta_{0})$)
If $n \rightarrow \infty$, then the probability tends to one.
I don't believe this is right. The $\hat p_{n}$ should be skewed towards $0$ because p values are likelier to be significant, but I am not sure where I am wrong. Help would be appreciated.