7
$\begingroup$

I'm tasked with deriving the MGF of a $\chi^2$ random variable.

I think the way to do is is by using the fact that $\Sigma_{j=1}^{m} Z^2_j$ is a $\chi^2$ R.V. and that MGF of a sum is the product of the MGFs of the individual terms. Although that may not be right and it may be $E(e^{tX})$ way.

I don't need it solved really just need to get down the track a little further.

$\endgroup$
1
9
$\begingroup$

Yes, since $\chi^2$ is a sum of $Z_i^2$ the MGF is a product of individual summands. But then you need the MGF of $Z_i^2$ which is $\chi^2$ with 1 degree of freedom. The obvious way of calculating the MGF of $\chi^2$ is by integrating. It is not that hard:

$$Ee^{tX}=\frac{1}{2^{k/2}\Gamma(k/2)}\int_0^\infty x^{k/2-1}e^{-x(1/2-t)}dx$$

Now do the change of variables $y=x(1/2-t)$, then note that you get Gamma function and the result is yours. If you want deeper insights (if there are any) try asking at http://math.stackexchange.com.

$\endgroup$
0
6
$\begingroup$

I think the easiest way is to simply start with a single squared gaussian: $$E[e^{tX^2}] = \int_{-\infty}^\infty e^{tx^2}\tfrac1{\sqrt{2\pi}}e^{-x^2/2}dx = \tfrac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(1-2t)x^2/2}dx = \tfrac1{\sqrt{1-2t}},$$ for $t<1/2$. Since the Chi-squared is just a sum of independent squared gaussians, you get the factor $k$ in the exponent.

$\endgroup$
1
  • 3
    $\begingroup$ +1 It's nice to see a clever approach that might even generalize to other situations. Of course this answer is limited to integral degrees of freedom. $\endgroup$
    – whuber
    Jul 13 '17 at 14:30
3
$\begingroup$

You can also do this calculation by brute-force straight from the general chi-squared distribution, without appeal to any intermediate appeal to sums of random variables. For $X \sim \chi_n^2$ we have moment generating function:

$$\begin{equation} \begin{aligned} M_X(t) \equiv \mathbb{E}(\exp (tX)) &= \int \limits_0^\infty \exp(tx) \cdot \text{Chi-Sq}(x | n) dx \\[8pt] &= \frac{1}{2^{n/2} \Gamma(n/2)} \int \limits_0^\infty \exp(tx) \cdot x^{n/2-1} \exp(-x/2) dx \\[8pt] &= \frac{1}{2^{n/2} \Gamma(n/2)} \int \limits_0^\infty x^{n/2-1} \exp((t -\tfrac{1}{2})x) dx. \\[8pt] \end{aligned} \end{equation}$$

For the case where $t < \tfrac{1}{2}$, using the change-of-variable $r = (\tfrac{1}{2} - t)x$ we have:

$$\begin{equation} \begin{aligned} M_X(t) &= \frac{1}{2^{n/2} \Gamma(n/2)} \int \limits_0^\infty x^{n/2-1} \exp((t -\tfrac{1}{2})x) dx. \\[8pt] &= (\tfrac{1}{2} - t)^{-n/2} \cdot \frac{1}{2^{n/2} \Gamma(n/2)} \int \limits_0^\infty r^{n/2-1} \exp(-r) dr. \\[8pt] &= (1 - 2t)^{-n/2} \cdot \frac{1}{\Gamma(n/2)} \int \limits_0^\infty r^{n/2-1} \exp(-r) dr. \\[8pt] &= (1 - 2t)^{-n/2}. \\[8pt] \end{aligned} \end{equation}$$

$\endgroup$
1
  • $\begingroup$ The details are nicely laid out, but the approach appears identical to that of the accepted answer. $\endgroup$
    – whuber
    Oct 30 '18 at 13:34

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.