Let $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $2$ and variance $2$. If $X_1$ and $X_2$ are Independent random variable then $P(X_1<X_2)$
My input
$\operatorname{Exp}(a)=ae^{-ax} \ \ \ ; G(a,\lambda)=\dfrac{a^{\lambda}}{\Gamma{\lambda}\ } e^{ax}x^{\lambda-1} $
$X_1\sim \operatorname{Exp}(1)=e^{-x}\implies G(1,1)$
Mean $ =\dfrac{\lambda}{a}=2 \implies \lambda=2a $
Variance = $\dfrac{\lambda}{a^2}=\dfrac{2a}{a^2}=2 \implies a=1,\lambda=2$
$X_2\sim G(1,2)$
$P(X_2-X_1<0)$
I am trying to find out distribution of $X_2-X_1$ I tried to subtract the parameter($\lambda_1,\lambda_2$) but the property of MGF works in addition only.
Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $X_2$ so I am out of option I need a hint or something.
Best Answer
Smiley is right.
Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1\geq X_2)=1-\int_0^\infty P(X_1\geq x)f_{X_2}(x) \, dx=1-\int_0^\infty e^{-x}f_{X_2}(x) \, dx$$
(and leave $X_2-X_1$ for what it is!)