# Proof required: Sum of squared standard normal random variables is a Chi-square rv

1. Oct 14, 2008

### stattheory

If Z1,Z2....Zn are standard normal random variable that are identically and independently distrubuted, then how can one prove that squaring and summing them will produce a Chi-
squared random variable with n degrees of freedom.

Any help on this will be greatly appreciated. I am new to this stuff and often get confused in it.

Stattheory.

2. Oct 15, 2008

### statdad

Re: Proof required: Sum of squared standard normal random variables is a Chi-square r

Several ways. Start with one, set up

$$F(a) = \Pr(X^2 \le a) = \Pr(-\sqrt{a} \le X \le \sqrt{a})$$

and show that $$f(a) = F'(a)$$ is the density for a central chi-square with 1 degree of freedom.

Then use the moment=generating method to show that the sum of $$n$$ chi-squares has a chi-square distribution with $$n$$ degrees of freedom.

Or, if you haven't seen moment-generating functions, start as above for 1, then
show that if $$W, Y$$ are two independent chi-square random variables, with $$n_1$$ and $$n_2$$ degrees of freedom, the sum $$W + Y$$ is chi-square with $$n_1 + n_2$$ degrees of freedom. The use induction for your case.

There are other ways, and I'm sure they will get proposed.

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