# Variance of $\chi^2$ statistic depends on $df$ but not on $n$?

I have an $$r\times c$$ contingency table. The chi-square statistic for the table is equal to $$\chi^2 = \sum_{i=1}^{rc} (O_i - E_i)^2 / E_i$$, where $$O_{i}$$ are the observed and expected counts for cell $$i$$ respectively. Let $$n$$ be the total number of counts. Then the variance of $$\chi^2$$ is given by $$2 \times (r-1) \times (c-1)$$, where $$(r-1) \times (c-1)$$ is the number of degrees of freedom. My question is the following – why does the variance not depend only on the $$df$$ and not on the total number of counts, $$n$$? It would seem to me that if $$n$$ is smaller there would be greater variability in the observed counts. If $$n$$ is extremely large, the observed counts would be very precise, leading to a more precise $$\chi^2$$ statistic.

• You need to pay attention to the information that went into computing the expected values $E_i.$ These issues are extensively discussed in other threads. Use keywords like "degrees of freedom" to search our site.
– whuber
Jul 10, 2019 at 16:09
• Sure the expected values $E_{i} = np_{i}$ and so depend on $n$. But the issue remains that if $n$ is larger both $O_{i}$ and $E_{i}$ are more precise and so $Var(\chi^2)$ should still intuitively depend on $n$. Jul 10, 2019 at 16:33
• Then it's important to modify your intuition: searching our site can help with that. Your premise is questionable, though: the statistic already accounts for the precision in the $O_i$ in dividing by $E_i,$ which is a (close) proxy for their variance.
– whuber
Jul 10, 2019 at 16:46
• "$rc$" is a misleading choice for the d.f in a chi-squared on a contingency table, as $r$ and $c$ conventionally stand for the number of rows and columns of the contingency table respectively, but then it is not the case that the df are $r \times c$ (almost never; it's actually somewhat tricky to come up with a realistic situation where you have that; if you are in such a situation, it might be worth mentioning it). Most typically the d.f. parameter in a test of independence is $(r-1)\times (c-1)$. . Jul 11, 2019 at 2:01
• @whuber Thanks, I see what you mean. It is the division by $E_i$ which normalizes the distribution. I suppose a simple analogy would be the z-statistic (for a Normal variate) which has fixed variance of 1 regardless of sample size, due to the standard error (of the Normal variate) in its denominator. Jul 14, 2019 at 15:17

Suppose $$X$$ has a binomial distribution with parameters $$n$$ and $$p$$, so with mean $$np$$, and variance $$np(1-p)$$ and fourth moment about the mean of $$3n^2p^2(1-p)^2-6np^2(1-p)^2-np(1-p)$$.
Then consider $$Y = \frac{(X-\mathbb E[X])^2}{\mathbb E[X]}$$. You can show that $$Y$$ has mean $$\frac{np(1-p)}{np}=1-p$$ which does not depend on $$n$$ and $$Y$$ has variance $$2(1-p)^2 -\frac{6}{n}(1-p)^2+\frac{1}{n}\frac{1-p}{p}$$; as $$n$$ increases, the variance of $$Y$$ tends towards $$2(1-p)^2$$, which also does not depend on $$n$$.
Now coming back to your question, one $${(O_i-E_i)^2/E_i}$$ is rather similar to the binomial case albeit affected by the row and column constraints, and similarly $$\sum_{i=1}^{rc} (O_{i} – E_{i})^{2} / E_{i}$$ is similar to the sum of an constant number of these though with some small covariances. So it is reasonable that, as $$n$$ increases, the variance of your $$\chi^2$$ heads towards a finite positive value which does not depend substantially on $$n$$.
In simpler form, $${(X-\mathbb E[X])^2}$$ has a mean which is $$O(n)$$ and a variance which is $$O(n^2)$$ so $$Y$$ has a mean and a variance which are both $$O(1)$$. Similarly $$(O_i-E_i)^2$$ has a mean which is $$O(n)$$ and a variance which is $$O(n^2)$$ so $${(O_i-E_i)^2/E_i}$$ and $$\sum_{i=1}^{rc} (O_{i} – E_{i})^{2} / E_{i}$$ have means and variances which are $$O(1)$$.
It is the division by $$\mathbb E[X]$$ and by $$E_i$$ which largely removes the effect of $$n$$ both on both the means and the variances, at least when $$n$$ is large.