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How would you find the probability of the following: $$\frac{c+\sum_{i=1}^na_iZ_i - \sum_{i=1}^ma'_iW_i}{\sqrt{\sum_{i=1}^nZ_i^2}}>\frac{d+\sum_{i=2}^{n+1}b_iZ_i - \sum_{i=1}^{m}b'_iW_i-b'_{m+1}Z_1}{\sqrt{\sum_{i=2}^{n+1}Z_i^2}}$$ where $Z_i \sim N(0,\sigma^2)$ for $1 \leq i \leq n+1$, $W_i \sim N(0,\sigma'^2)$.

I tried to square and multiply through by the denominators and then tried to see if I could combine it into one gamma distribution, but I got to a point of trying to combine dependent gamma distributions and am not sure where to go from there.

If anyone has a suggestion for a mathematical or computational approach for solving this that would be a great help.

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    $\begingroup$ How large are $m$ and $n$? $\endgroup$
    – user225256
    Commented Sep 29, 2023 at 12:38
  • $\begingroup$ n is around 10, and m could range from 10 to 100. $\endgroup$
    – mrepic1123
    Commented Sep 29, 2023 at 12:45

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This is not an answer, but how I would think about this:

Let $Q,R,S,T$ be standard (multi)normal variables of dimensions $1,n-1,1,m$.

Let $e=\sigma a_1$, $f=\sigma a_{2..n}$, $g=\sigma b_{2..n}$ and $h=\sigma b_{n+1}$ (of dimensions $1,n-1,n-1,1$).

Then this becomes $$\frac{ c+eQ+f.R-\sigma’a’.T }{ \sigma\sqrt{Q^2+R.R} }>\frac { d+g.R+hS-\sigma’b’.T-\sigma b’_{m+1}Q }{ \sigma\sqrt{R.R+S^2} }$$

So this can be phrased as a question about:

  • six standard normal variables ($Q$, $S$, two components of $R$ and two components of $T$)
  • and one chi-square variable, for the remaining $n-3$ degrees of freedom of $R.R$.

That can probably be reduced further, but even so it looks too ugly to calculate well.

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