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.
 A: Let's start with a different question, and then argue by analogy:
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.
