Minimum between gamma and exp

Question

For $X_1,X_2 \sim Exp(\gamma_i)$

it is known that $ Min\{X_1,X_2\} \sim Exp(\gamma_1 + \gamma_2) $

and $ \Pr(X_1 < X_2) = \dfrac{\gamma_1}{\gamma_1+\gamma_2} $

Can we say something similar when one of the variables is distributed gamma?

For $X \sim Exp(\gamma_1)$ and $Y \sim Gamma(\alpha,\gamma_2)$

$ \Pr(Y < X) =? $

is it true that $ \Pr(Y < X) = \left( \dfrac{\gamma_2}{\gamma_1+\gamma_2} \right)^{\alpha} $

Answer 1

Almost so, just need to change the indices. ${\text Pr}(Y

Indeed, $$ {\text Pr}(Yy)dy = \int_0^\infty \dfrac{\gamma_2^\alpha}{\Gamma(\alpha)}y^{\alpha-1}e^{-\gamma_2y} e^{-\gamma_1y}dy = \left(\dfrac{\gamma_2}{\gamma_1+\gamma_2}\right)^\alpha. $$

One cannot see the similar results for minimum between exponential and gamma. The CDF of minimum is $$ {\text Pr}(\min(X,Y)\leq x) = 1-(1-F_X(x))(1-F_Y(x))=1-e^{-\gamma_1x}(1-F_Y(x)). $$ Take $\alpha=2$ for example. Then $1-F_Y(x)=e^{-\gamma_2x}(1+\gamma_2x)$, and $$ {\text Pr}(\min(X,Y)\leq x) = 1-e^{-(\gamma_1+\gamma_2)x}(1+\gamma_2x),$$ which is not CDF of gamma distribution.