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


1 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 $$ \phi(x-z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-z)^2}{2}} \implies \phi'(x-z) = (x-z) \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-z)^2}{2}}= (x-z) \phi(x-z) $$

  • $\begingroup$ Thank you very much! Could you please explain how you solved for the derivative of the Gaussian CDF wrt z? $\endgroup$
    – taurus
    Oct 30, 2020 at 5:00
  • $\begingroup$ please see the edit $\endgroup$ Oct 30, 2020 at 5:14
  • $\begingroup$ 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? $\endgroup$
    – taurus
    Oct 30, 2020 at 5:23
  • 1
    $\begingroup$ 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: $\frac{d}{dz} \Phi(x-z) = -\phi(x-z)$ $\endgroup$ Oct 30, 2020 at 16:11
  • $\begingroup$ Ah noted. I won't make that mistake again. Thanks. $\endgroup$
    – taurus
    Oct 30, 2020 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.