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What is the probability of the following: $P\left(Z_j>\frac{\epsilon}{a_j*R}\left(\sum^{M}_{m=j+1} ~ a_m*R*X_m+1\right)\right)$

where $Z_j$ and $X_m$ are independent and identically distributed random variables, where they are continuous random variables that take values from 0 to infinity. having the following CDF:

$F_{Z}(z){=}\frac{\gamma\left(B,\frac{\sqrt{z}} {u}\right)}{\Gamma(B)}$

$F_{X_m}(x){=}\frac{\gamma\left(B,\frac{\sqrt{x}} {u}\right)}{\Gamma(B)}$

$a_m$, $\epsilon$, $u$, and $R$ are constants.

The summation of all $a_m$ is: $\sum^{M}_{m=1}a_m=1$ where $M=6$.

learning statistics
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  • Please don't repost a question when it's closed: just edit it. In this instance, continuing my comment to your now deleted first version, the difference of random variables appears to be a linear combination of Gamma variates (if I guess correctly about your notation). See https://stats.stackexchange.com/questions/72479 for a general method of computing its distributional properties. – whuber Nov 16 '23 at 14:09
  • so the first step I should move the right hand side to the left hand side like this $P\left(Z_j-\frac{\epsilon}{a_jR}\sum^{M}_{m=j+1} ~ a_mR*X_m>1\right)$ – learning statistics Nov 16 '23 at 16:30
  • The idea is right but the algebra is incorrect. – whuber Nov 16 '23 at 16:41
  • Thank you for your answer I mean this $P\left(Z_j-\frac{\epsilon}{a_jR}\sum^{M}_{m=j+1} ~ a_mRX_m>\frac{\epsilon}{a_jR}\right)$ – learning statistics Nov 16 '23 at 16:45
  • That looks better ;-). – whuber Nov 16 '23 at 19:57

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