I've got difficulties in computing the second momentum of chi squared.

Chi squared distribution with $n$ degrees of freedom is the sum of $n$ independent distributions $X^2$, where $X \sim N(0;1)$.

We know that the fourth momentum of each $X_i$ is $3$.

So, mark $\chi^2 = C$, and $$E(C^2) = E\left[ \left(\sum X^2\right)^2 \right] = \sum E\left[X^4\right] = 3n$$ where the penultimate equality is due to $X$s being iid and linearity of E-operator.

The correct answer is $n^2 +2n$, so there is something wrong with the above calculation, but I just can't spot what it is. (Please note that I'm not claiming that $\left(\sum x^2 \right)^ 2 = \sum x^4$ in general. Here it should hold however, since the $X$s are independent, right?)

  • $\begingroup$ I have edited your question to make use of $\LaTeX$ math formatting. Please confirm that I have not misrepresented your question by editing. :) $\endgroup$ – apnorton Jul 17 '14 at 15:19
  • $\begingroup$ Yeah, thanks and sorry. No misrepresentation there! :) $\endgroup$ – user3010768 Jul 17 '14 at 15:24
  • $\begingroup$ That's good to hear. And no need to apologize--LaTeX/MathJax can take a bit to get used to. Here's the "quick reference" guide for this site. (Unfortunately, I don't know statistics, so I can't help with your question...) $\endgroup$ – apnorton Jul 17 '14 at 15:26
  • $\begingroup$ Independence would suggest $E[X_i^2 X_j^2] = E[X_i^2]E[X_j^2]$ for $i \not = j$ but that is unlikely to be zero. $\endgroup$ – Henry Jul 17 '14 at 15:36


$E\left[ \left(\sum_i X_i^2\right)^2 \right] = E\left[ \sum_i X_i^4 \right] + E\left[ \sum_i \sum_{j \not = i}X_i^2 X_j^2 \right] = 3n+n(n-1)$

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Your argument in parentheses is wrong. Rather, we always have, regardless of independence, that

$$(\sum X^2)^2=\sum_i X_i^4+2\sum_{i<j}X^2_iX^2_j$$

The independence helps when we take expectations. Since the X's are independent we have $$E[X_iX_j]=E[X_i]E[X_j]$$

We also have that because the X's are independent, so are their squares. (See Are squares of independent random variables independent?.) Therefore, combining the two previous facts, taking expectations of the above sum gives $$3n+2\frac{n(n-1)}{2}=3n+n^2-n=n^2+2n$$

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  • $\begingroup$ Ok, now I got it! I misremembered independence to mean that E[XY] = 0. Thanks for your answers. :) $\endgroup$ – user3010768 Jul 17 '14 at 15:40
  • $\begingroup$ You're welcome. Please consider upvoting them. $\endgroup$ – symplectomorphic Jul 17 '14 at 15:47

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