Finding the power function and show its convergence Let $X_1, \ldots , X_n \sim N(\theta, 1).$ Consider testing $H_0: \theta = 0$ versus $H_1: \theta = 1$.
Let the rejection region be $R = \{x^n : T(x^n) > c\}$ where $T(x^n) = n^{-1}\sum^n_{i = 1} X_i$.
a) Find $c$ so that the test has size $\alpha$.
b) Find the power under $H_1$, that is, find $\beta(1)$.
c) Show that $\beta(1) \to 1$ as $n \to \infty$.
Right now I've been stuck on part b, because the answer I get converges to $0$ rather than $1$ (for large $n$). The power function I got (using Central Limit Theorem) is $\beta(\theta) = P_\theta(T(x^n) > c) = P(Z_n > \sqrt{n}(c-\theta)) = 1 - \phi(\sqrt{n} (c-\theta))$, where $Z_n$ has a standard normal distribution and $\phi$ is the respective CDF. Supposedly, subsituting $\theta = 1$ would give the answer to b), but I feel this is wrong since this doesn't converge to 1 for large n.
If someone could help point out my error, that would be much appreciated. Thank you kindly in advance!
 A: A.
Need to find $c$ such that $\alpha = P[\bar{X}>c | \theta = 0]$.
Note that $\bar{X} \sim N(\theta,\frac{1}{n})$, and that conditioned on $\theta=0$ this implies that $\bar{X} \sim N(0,\frac{1}{n})$.
So we need to find $c$ such that $.05 = P[N(0,\frac{1}{n})>c]$. This can be re-arranged to be: $.05 = P[N(0,1)>\sqrt(n)c]$. Getting this in the form of the CDF, we express it as:
\begin{equation}
.95 = P[N(0,1) \leq \sqrt{n}c] \Leftrightarrow \sqrt{n}c = \Phi^{-1}(.95) \Leftrightarrow c = \frac{\Phi^{-1}(.95)}{\sqrt{n}}
\end{equation}
B. 
We have that $\beta(1) = P[\bar{X}>c | \theta = 1]$. Again in this setting we have a closed form for $\bar{X}$ conditional on $\theta = 1$: $\bar{X} \sim N(1,\frac{1}{n})$. 
(Note here that we aren't using the central limit theorem, just the fact that normal distributions are closed under linear combinations.)
Plugging in our value for $c$ gives us:
\begin{equation}
\begin{split}
P[\bar{X}>c | \theta = 1] &= P[N(1,\frac{1}{n})>\frac{\Phi^{-1}(.95)}{\sqrt{n}}]\\
& = P[N(\sqrt{n},1)>\Phi^{-1}(.95)]\\
& = P[N(0,1)>\Phi^{-1}(.95) - \sqrt{n}]\\
\end{split}
\end{equation}
Using the symmetry of the standard normal around 0, we get:
\begin{equation}
P[N(0,1)>\Phi^{-1}(.95) - \sqrt{n}] = P[N(0,1)\leq \sqrt{n}-\Phi^{-1}(.95)] = \Phi(\sqrt{n} - \Phi^{-1}(.95))
\end{equation}
C. 
We can see that because $\lim_{x\rightarrow\infty} \Phi(x) = 1$ that we reach power of 1 asymptotically as $n\rightarrow \infty$.
Edit It's also kind of interesting to note that the rate at which the power increases slows down for multiple reasons as $n$ gets larger. Both $\Phi(x)$ and $\sqrt{x}$ have negative second derivatives for large $x$:
\begin{equation}
D^2 \sqrt{x} = \frac{-1}{4x^{3/2}}
\end{equation}
\begin{equation}
D^2 \Phi(x) = \frac{-x}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}
\end{equation}
Which means that you are both approaching infinity slower, and also slowing down the pace at which an increase in $x$ put into $\Phi(x)$ increases your power.
That wasn't really related to the question, I just thought it was cool.
