I am looking at a paper that is interested in finding a probability that one chi-squared variable is smaller than another.
More precisely
$$ \Pr\left[\sum_{i=1}^k\left(X_i-Y_i+Z_i\right)^2\leq \sum_{i=1}^k\left(Z_i\right)^2\right] \\ X,Y\sim\mathcal{N}(0,\sigma^2) \text{ and } Z\sim\mathcal{N}(0,1) $$ and all three RV are independent.
the paper states that that this probability is
$$ = 1-\mathrm{E}_\theta\left\{\mathcal{Q}_{k/2}\left(\sqrt{\frac{\theta}{2\sigma^2}}, \sqrt{\frac{\theta}{2\sigma^2}}\right)\right\} $$
Where $\mathcal{Q}$ is the Marcum Q Function and $\theta\sim\chi_k^2$.
Running trials this does seem to be correct, but I don't know enough about the Marcum Q function to see how they derived this.
I am guessing this is related to the Marcum Q expression for the CDF of a non central chi-square distribution.
Any pointers are greatly appreciated.
Update I think I figured out the broad strokes.... if we work through the equation then we get something like $$ \Pr\left[\sum_{i=1}^k\left(X_i^2 +Y_i^2+Z_i^2-2X_iY_i+2X_iN_i-2Y_iN_i\right)\leq \sum_{i=1}^k\left(Z_i\right)^2\right] \Pr\left[\sum_{i=1}^k-\left(X_i^2 -2X_iY_i+2X_iN_i-2Y_iN_i\right)\geq \sum_{i=1}^k\left(Y_i\right)^2\right] $$
which seems to make some sense as we take the expectation over all possible $Y$ we get the integration limit in $\mathcal{Q}$. I will have to see if I can coax the left side to be a non-central chi-squared distribution with the correct $\lambda$ non-centrality parameter as it's evaluated for fixed Y_i.