# Why does Pearson's chi-squared test divide by the mean and not the variance?

I am wondering why in Pearson's chi-squared test, the divisor of each element in the sum is the matching expectation and not the matching variance.

As I understand it, the test works by standardizing each normal variable before summing, so the results set can be tested against the chi-squared distribution which deals with a sum of squares of standard normal random variables.

The way a normal random variable is standardized is by subtracting the expectation and dividing by the standard deviation. So, in Pearson's test, this should give the variance in the divisor of each element, not the expectation.

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Notice that $\{O_i\}$ are not independent, their total must equal 1. More careful analysis must be made to establish the law of the statistics. –  Sasha Dec 6 '12 at 16:56

## 1 Answer

Intuition/informal proof: The expected value is equal to the variance, so when you divide by the expected value you are in fact dividing by the variance, as you thought you should. If you think of it in terms of counts that follow a Poisson distribution this is natural, since the mean and variance of a $\operatorname{Poisson}(\lambda)$ distribution are both $\lambda$.

For a formal proof, check out MIT's OpenCourseWare.

Great question!

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Thank you for the reference to the proof! Could you please elaborate on the intuition? The Poisson distribution is what the Binomial distribution tends to as n->inf and p->0. This is not the general case in Pearson's test. –  Danra Dec 7 '12 at 1:38
The Poisson distribution has an independent existence, not just as the limit of the Binomial distribution. It's the distribution of the number of "rare" events that occur in a given time period, and is very often used to model counts. In each cell of the contingency table we are counting how many observations fall into that cell, so you can think of the counts in the contingency table like Poisson-distributed random variables. It's not formal--in most cases those cells aren't actually Poisson-distributed, for various reasons--but I think it's a reasonable intuitive argument. –  Jonathan Christensen Dec 7 '12 at 1:54
As I understand it, a Poisson distribution is appropriate when n is large and p is small; See en.wikipedia.org/wiki/… . Granted, "large" and "small" are relative terms; But does the Poisson-ic intuition really hold, for instance, in the case of a repeated coin toss, where there are only two cells, and p0 and p1 are both "large"? Isn't it more intuitive that the variances in this case are of binomial distributions? –  Danra Dec 7 '12 at 8:55
It's intuition. I find it useful because I often think of counts like those found in contingency tables in terms of Poisson random variables. If you don't find it useful then ignore it. If you want a rigorous answer, work through the rigorous proof. –  Jonathan Christensen Dec 7 '12 at 17:16