If $X\sim \operatorname{Exp}(\lambda_1)$ and $Y\sim \operatorname{Exp}(\lambda_2)$ are independent RVs. Find the following:
a) $ P(X< Y) $
I tried finding this with $\int_{-\infty}^{\infty} F_X(y)f_y(y)\,dy$, but the integral calculator couldn't do it.
b) Distribution of the $\operatorname{Min}(X,Y)$ (hint : think of the CDF)
I have no idea what this means. The distribution of the minimum between the 2?
c) Distribution of $X + Y$
I'm assuming this question means the closest approximate distribution to their sum?
We compute $\mathbb P(X<Y)$ from the joint density: \begin{align} \mathbb P(X<Y) &= \int_{\mathbb R^2}f_{X,Y}\mathsf 1_{\{X<Y\}}\\ &= \int_0^\infty \int_0^y f_{X,Y}(x,y)\ \mathsf dx\ \mathsf dy\\ &= \int_0^\infty \lambda_2 e^{-\lambda_2y} \int_0^y \lambda_1e^{-\lambda_1 x}\ \mathsf dx \ \mathsf dy\\ &= \int_0^\infty \lambda_2e^{-\lambda_2y}(1 -e^{-\lambda_1 y} )\ \mathsf dy\\ &= \int_0^\infty \lambda_2 e^{-\lambda_2y}\ \mathsf dy - \lambda_2\int_0^\infty e^{-(\lambda_1+\lambda_2)y}\ \mathsf dy\\ &= 1 - \frac{\lambda_2}{\lambda_1+\lambda_2}\\ &= \frac{\lambda_1}{\lambda_1+\lambda_2}. \end{align} Let $Z=X\wedge Y$. Then for any $t>0$, $$ \mathbb P(Z>t) = \mathbb P(X>t, Y>t) = \mathbb P(X>t)\mathbb P(Y>t) = e^{-\lambda_1t}e^{-\lambda_2t}=e^{-(\lambda_1+\lambda_2)t},$$ so that $Z$ has exponential distribution with parameter $\lambda_1+\lambda_2$.
To compute the density of $W=X+Y$, we use convolution. For any $t>0$ we have \begin{align} f_W(t) &= (f_X\star f_Y)(t)\\ &= \int_{\mathbb R}f_X(\tau)f_Y(t-\tau)\ \mathsf d\tau\\ &= \int_0^t \lambda_1 e^{-\lambda_1\tau}\lambda_2 e^{-\lambda_2(t-\tau)}\ \mathsf d\tau\\ &= \lambda_1\lambda_2 e^{-\lambda_2 t}\int_0^t e^{-(\lambda_1-\lambda_2)\tau}\ \mathsf d\tau\\ &= \frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2}e^{-\lambda_2 t}(1 - e^{-(\lambda_1-\lambda_2) t}).\\ &= \frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2}(e^{-\lambda_2 t}-e^{-\lambda_1 t}), \end{align} assuming $\lambda_1\ne\lambda_2$. If $\lambda_1=\lambda_2=\lambda$ then we have \begin{align} (f_X\star f_Y)(t) &= \int_0^t \lambda^2 e^{-\lambda \tau}e^{-\lambda(t-\tau)}\ \mathsf d\tau\\ &= \lambda^2 e^{-\lambda t} \int_0^t \mathsf d\tau\\ &= \lambda^2 t e^{-\lambda t}. \end{align}