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In an attempt to derive something, I have come to a point where I need to know what the following can be simplified to. It looks to me that I am looking for the 4th moment of ''y''. I am just not sure how to go about this.

$$E\space[\space(\space\sum_{i=1}^N (\space y_i^2\space)\space)^2\space]= \space ?$$

For my problem, the following relationship is valid$$ E\space[\sum_{i=1}^N\space y_i^2 \space] = \space N\sigma^2 $$

Note that: $$y_i \sim \space N(0,\sigma^2) $$

user511482
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1 Answers1

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It's not a fourth moment it's actually a second (non-central) moment of the chi square.

Up to a constant the sum $\sum(y_i^2)$ is a chi-squared random variable with $N$ degress of freedom. So up to a constant this is the second non-central moment of a chi square with $N$ degrees of freedom.

Look here for the formula for the second non-central moment: https://en.wikipedia.org/wiki/Chi-squared_distribution#Noncentral_moments

Gregory Grant
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  • I should make the correction. The sum runs from 1 to N. Not 0 to N. So there are N number of random variables yi. And they are all independent. So now that I have made the correction, the sum runs from 1 to N. So it is indeed $$ N \sigma^2 $$ as mentioned.

    Thank you for your response. I will have to look more into chi-squared r.v. before I can get to a conclusion

     – user511482 Dec 08 '17 at 21:20
  • Thank you for your response. When you say $$ X = \sum Y_i^2 $$ then E[X] is not 0. It is as I said, $$ N\sigma^2 $$ – user511482 Dec 09 '17 at 02:05
  • Yes sorry, it's the second non-central moment of a chi-squared, there's a formula for that on the chi square wiki page. I updated my answer to fix the mistake. – Gregory Grant Dec 09 '17 at 09:47
  • Based on the wikipedia page, the non-central moment of a chi-squared random variable with k degrees of freedom is:

    $$ E[X^m] \space = \space k(k+2)(k+4)...(k+2m-2) $$

    We are interested in second non-central moment of the chi-square which means m=2 in the above equation and since our sum runs from 1 to N, there are N degrees of freedom, just as you suggested. This makes k=N.

    The above expression reduces down to:

    $$ E[X^2] \space = \space N(N+2) $$

    The original question that was asked therefore equals:

    $$E[(\space\sum_{i=1}^N \space(y_i^2))^2\space]= \space N(N+2)$$

     – user511482 Dec 09 '17 at 18:00
  • Well no be careful because $\sum y_i^2$ only equals a chi square up to a constant. You have to divide by $\sigma^2$ or something like that to get each of the $y_i^2$ a standard normal. – Gregory Grant Dec 09 '17 at 22:17
  • Well, I can't seem to figure out exactly what that something is that I need to divide. – user511482 Dec 11 '17 at 17:27