I am studying for a test, one section of which will cover the delta method. This problem came from that section:

Let $X\sim N(\mu,n^{-1})$. Find an approximate distribution of $X^2$. (It also asks for exact distribution, but I can do that).

Can you help me get started? I can't see how I can apply delta method or central limit theorem etc. to find some approximation.

I know that $\sqrt{n}(X-\mu) \sim N(0,1)$, but I don't see how that can help either.


Let's double check what the delta method is:

Roughly, if there is a sequence of random variables Xn satisfying $$ {\sqrt{n}[X_n-\theta]\,\xrightarrow{D}\,\mathcal{N}(0,\sigma^2)}, $$ where $\theta$ and $\sigma^2$ are finite valued constants and $\xrightarrow{D}$ denotes convergence in distribution, then $$ {\sqrt{n}[g(X_n)-g(\theta)]\,\xrightarrow{D}\,\mathcal{N}(0,\sigma^2[g'(\theta)]^2)} $$ for any function $g$ satisfying the property that $g′(\theta)$ exists and is non-zero valued.

Okay, so showing you have the first thing already is easy right?

So let $g(x) = x^2$ and you're set, aren't you? Just apply the theorem.

Then at the end do the relevant linear transformation (affecting variance and mean) so that you're just talking about what $g(X)$ will approximately be distributed as.

So in short:

i) State the theorem.

ii) Explain/show how the first part is okay

iii) state $g$

iv) apply the theorem

v) infer approximate distribution for $g(X)$

Which is pretty much what you do any time you want to use it.

  • $\begingroup$ Well I $\sqrt{n}(X-\mu) \sim N(0,1)$ which is a bit different, but I suppose it is stronger than simply approaching in distribution... $\endgroup$
    – bdeonovic
    Feb 21 '13 at 3:08
  • $\begingroup$ Since it's already there, it gets there really fast. $\endgroup$
    – Glen_b
    Feb 21 '13 at 3:10
  • $\begingroup$ Yes, I see. I guess it was a silly question then, also quite easy :) $\endgroup$
    – bdeonovic
    Feb 21 '13 at 3:12
  • $\begingroup$ Actually this is kind of an interesting problem. I get something like $X^2 \overset{\circ}{\sim} N(\mu^2,4\mu^2/n)$ which is fine as long as $\mu\ne 0$... perhaps using a second order taylor expansion in delta method will be better. $\endgroup$
    – bdeonovic
    Feb 21 '13 at 3:17
  • 1
    $\begingroup$ aha! Turns out a better approximation when $\mu=0$ is a $\chi_1^2$ random variable, which makes perfect sense since a normal random variable squared is something like a $\chi^2$ random variable. $\endgroup$
    – bdeonovic
    Feb 21 '13 at 3:37

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