Let $X_1, X_2, \ldots X_n$ be independent Gaussian random variables with $N(0,1)$ distribution. Find the density and the characteristic function of $X_1^2 + X_2^2 + \cdots + X_n^2$

I am preparing for exams and I am stuck with this particular problem. I have seen the derivation for the characteristic function of a standard normal random variable. But I tried calculating the integral for the square and I cannot get it. Once I do that though, I think I can just raise my answer to the nth power to account for the sum. Can someone please help me with this? I have been stuck and not sure for many hours.


Let $Y=\sum_{k=1}^n X_k^2$. The char. function of $Y$ is $\phi(t)=E(e^{itY})=\prod_{k=1}^nE(e^{itX_k^2})$, because of independence.

$E(e^{itX_k^2})=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{x^2}{2}+itx^2}dx=g(t)$. Therefore $\phi(t)=g^n(t)$.

I believe $g(t)=\frac{1}{\sqrt{1-2it}}$.

  • 1
    $\begingroup$ Thanks! How can I get the density from here? I am pretty sure it is a chi-square distribution. I am trying to do Fourier inversion $\endgroup$ – hom Dec 2 at 23:10
  • 1
    $\begingroup$ Should it be $g(t) = \frac{1}{\sqrt{1 - 2it}}$? $\endgroup$ – hom Dec 3 at 1:36
  • $\begingroup$ Yes - I got mixed up in the final step to get $g(t)$.. I corrected the answer. Getting the density function seems tough. Fourier inversion is correct, but it looks very difficult. $\endgroup$ – herb steinberg Dec 3 at 4:49
  • $\begingroup$ @hom I believe that the density function might be derivable by diirect integration. I did the first three $(Y_n=\sum_{k=1}^n X_k^2)$ and $f_n(y)$ are the density functions for$Y_n$ I got $f_1(y)=\frac{e^{-y/2}}{\sqrt{2\pi y}}$ , $f_2(y)=\frac{e^{-y/2}}{2}$, and $f_3(y)=e^{-y/2}\sqrt{\frac{y}{2\pi}}$ $\endgroup$ – herb steinberg Dec 5 at 1:39
  • $\begingroup$ @hom Since $f_2(y)$ is so simple, you can the rest of list by doing the odd and even numbered members as two lists by repeated convolutions using $f_2(y)$. $\endgroup$ – herb steinberg Dec 5 at 3:38

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.