Expectation of Gaussian CDF

Question

I have $X_i$ from $n$ i.i.d draws of N($\theta$,1). By the SLLN, I know $\bar{X_n}$ converges almost surely to $\theta$. I would like to find the limiting distribution of $\phi(x-\bar{X_n})$ for fixed $x$ where $\phi$ is the CDF of a standard normal. Is it valid to say $E\phi(x-\bar{X_n}) = \phi(x-E\bar{X_n}) = \phi(x-\theta)$ as $n$ approaches $\infty$? If not, how can I find the expectation of $\phi(x-\bar{X_n})$?

Answer 1

You need to use a combination of the Central Limit Theorem and the Delta method. By the CLT, we know that $$ \sqrt{n}(\bar{X}_n - \theta) \implies N(0,1), $$ where $\implies$ denotes convergence in distribution. Then, the delta method tells us that for any function $g$ such that $g'(\theta)$ exists and is not equal to zero, that $$ \sqrt{n}(g(\bar{X}_n) - g(\theta)) \implies N(0, [g'(\theta)]^2) $$ Now, for your case, you can take (for a fixed $x$): $$ g_x(z) = \phi(x-z), $$ which has first derivative with respect to $z$: $$ g'_x(z) = (x-z)\phi(x-z) $$ So we get $$ \sqrt{n}(\phi(x-\bar{X}_n) - \phi(x-\theta)) \implies N(0, (x-\theta)^2 \phi^2(x-\theta)) $$

An important thing to note here is that this breaks down for $x=\theta$, since the variance becomes zero and this is no longer a limiting distribution. To get around this, we can use the second order delta method. You can read more about all of this on the wiki page.

edit: To differentiate $\phi(x-z)$ with respect to $z$, we note that $$ \phi(x-z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-z)^2}{2}} \implies \phi'(x-z) = (x-z) \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-z)^2}{2}}= (x-z) \phi(x-z) $$