On page 181 of Handbook of Applied Cryptography Chapter Five, it states the following:
The purpose of this test is to determine whether the number of occurrences of 00, 01, 10, and 11 as subsequences of s are approximately the same, as would be expected for a random sequence. Let $n_0$, $n_1$ denote the number of 0’s and 1’s in s, respectively, and let $n_{00}$,$n_{01}$,$n_{10}$,$n_{11}$ denote the number of occurrences of 00, 01, 10, 11 in s, respectively. Note that $n_{00} + n_{01} + n_{10} + n_{11} = (n − 1)$ since the subsequences are allowed to overlap. The statistic used is $$ X_2 = \frac{4}{n-1}(n_{00}^2+n_{01}^2+n_{10}^2+n_{11}^2)-\frac2n (n_0^2 +n_1^2) +1$$
Where does this statistic come from? I'm not entirely sure, and the book doesn't explain it's derivation. Normally, $ X=\frac{(Z-\mu)^2}{n^2}$ for Chi-Squared test ( if i recall correctly), but I don't see how this can result in the expression above. For each $n_{ij}$, the approximate value is $\frac{n_i+n_j-1}{4}$, and the expected value would be $\frac{n_i+n_j}2$, but how do we combine them in such a way that we get the expression for $X_2$?
If this belongs in Math, rather than Cryptography, I can happily delete and move it. I just figured since the question is cryptography related, it may receive better attention here.