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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1$\begingroup$ How large are $m$ and $n$? $\endgroup$– user225256Commented Sep 29, 2023 at 12:38
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$\begingroup$ n is around 10, and m could range from 10 to 100. $\endgroup$– mrepic1123Commented Sep 29, 2023 at 12:45
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1 Answer
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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.