We are given $n$ i.i.d Gaussian random variables $x_i$ with mean $0$ and variance $1$. We construct two Chi-squared random variables $Y=\sum_{i \in I_1} x_i^2$ and $Z=\sum_{i \in I_2} x_i^2$ with $|I_1|=|I_2|$. Compute $\Pr(Y\le Z)$.
I know that if $Y$ and $Z$ are independent $A:=Y-Z$ would be a random variable as well that is Chi-square distributed. And then we could compute the probability that $\Pr(A\ge 0)$ with the regularized Gaussian function $P$.
But what do I do if they are not independent?
If $Y$ and $Z$ overlap over some $k$ terms, they disappear from the summation $Z-Y$. For example if $n=6$ we may have $I_1=\{1,2,3,4\}$ and $I_2=\{3,4,5,6\}$ so we would have $$Z-Y=(x_5^2+x_6^2)-(x_1^2+x_2^2)$$ Since $I_1$ and $I_2$ have the same length, their difference will always result in $m$ $x_j^2$'s minus another $m$ $x_j^2$'s where $m$ is given by $|I_1|-k=|I_2|-k$. This means $$Z-Y=X-W$$ where $X$ and $W$ are independent chisquared distributed with $m$ degrees of freedom.