Question

Loss amount, X, follows a gamma distribution. You are given E[X] = 8, Var[X] = 32. Calculate the probability that the loss amount exceeds 3, Pr[X>3].

Answer 1

$\text{For gamma distribution}, \\ E(x)=\frac{n}{\lambda}, V(x)=\frac{n}{\lambda^2},then,\\ \frac{n}{\lambda}=8........(i)\\ \frac{n}{\lambda^2}=32.....(ii)\\ \text{By dividing\;(i)by(ii)}\\ \therefore \lambda=0.25, n=2\\ f(x)=0.0625xe^{-0.25x}, x\geq 0\\ Pr[X>3]=\int^{\infty}_{3}0.0625xe^{-0.25x}dx\\ =0.0625[\frac{-xe^{-0.25x}}{0.25}-\frac{e^{-0.25x}}{0.0625}]^\infty_3\\ =0.0625[0-(\frac{-3e^{-0.25(3)}}{0.25}-\frac{e^{-0.25(3)}}{0.0625})]\\ \approx 0.827$