# Expectation of Gaussian CDF

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 $$\phi(x-\bar{X_n})$$?

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 $$\phi(x-z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-z)^2}{2}} \implies \phi'(x-z) = (x-z) \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-z)^2}{2}}= (x-z) \phi(x-z)$$
• Isn't CDF $\phi(x-z)$ defined as an integral from $-\infty$ to $x-z$? If so, isn't $\phi'(x-z)$ the standard normal PDF evaluated at $x-z$ since the derivative of CDF is a PDF? Oct 30, 2020 at 5:23
• oh, I misread your question, $\phi$ usually denotes the pdf of a standard normal, and $\Phi$ the cdf. Using this notation, we have that $\frac{d}{dz} \Phi(z) = \phi(z)$ i.e. the derivative of the CDF is the PDF, so in our case we have using the chain rule: $\frac{d}{dz} \Phi(x-z) = -\phi(x-z)$ Oct 30, 2020 at 16:11