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I have 60 numbers drawn from a normal distribution with mean 0 and standard deviation of 1.

1 realization:

enter image description here

I then take the sum of the 60 values.

I do this 1000 times and plot a histogram of the various sums I get:

enter image description here

And I am able to fit the histogram with a normal distribution with a mean of 0 and a standard deviation of sqrt(60).

So far so good.

I now want to square each of my 60 numbers to get (for example) a realization like this:

enter image description here

If I again take the sum of the 60 values and repeat 1000 times I get a different histogram:

enter image description here

My question: what function do I use to fit this histogram? Is it another normal distribution? Maybe chi-squared? What parameters do I use?

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  • $\begingroup$ What is the horizontal axis in your "realization" graphs? $\endgroup$
    – Acccumulation
    Commented Apr 16, 2023 at 1:07
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    $\begingroup$ I'm sure those are indices. It really doesn't matter. A point graph would be more appropriate than a line graph. The lines imply some connection between the points but the only connection between them is the negligible connection provided by the pseudo-random number generator. $\endgroup$
    – Joooeey
    Commented Apr 16, 2023 at 19:06

1 Answer 1

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If $z$ has a standard normal distribution (mean of zero and variance of one), then $z^2$ has a chi-squared distribution with 1 degree of freedom.

Furthermore, the sum of $x$ and $y$ from chi-squared distributions with degrees freedom $\nu_x$ and $\nu_y$ is itself chi-squared distributed (with $\nu_x+\nu_y$ degrees of freedom).

Thus, you are looking at a sampling distribution from a chi-squared distribution with 60 degrees of freedom. Furthermore, the chi-squared distribution approaches a normal distribution in the limit with mean equal to the degrees of freedom (and variance is double this value).

(Proof of these statements can be found in any undergraduate mathematical statistics and probably textbook...so I'll omit a formal proof here.)

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    $\begingroup$ You might want to mention the assumption of independence needed for summing squared standard normals (or chi-squares) to obtain a chi-squared distribution. $\endgroup$
    – Richard Hardy
    Commented Apr 15, 2023 at 16:29

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