I have $X\sim Exp\left(\lambda_{1}\right)$ and $Y\sim Exp\left(\lambda_{2}\right)$ that are Independence. How would i want to calculate $P\left(X>Y\right)$ ?
I know that $\int_{0}^{\infty}\lambda_{1}e^{-\lambda_{1}x}\cdot\int_{0}^{\infty}\lambda_{2}e^{-\lambda_{2}x} = \int_{0}^{\infty}\int_{0}^{\infty}\lambda_{1}\lambda_{2}e^{\left(-\lambda_{1}-\lambda_{2}\right)x}$
How this could help me ?
$$P(X>Y)=\int_0^{\infty}P(Y $$=\int_0^{\infty}P(Y Or if you want to work with the common density $$f_{X,Y}(x,y)=\lambda_1\lambda_2e^{-(\lambda_1x+\lambda_2y)}$$ for $x,y\geq 0$ then you have to integrate this common density over the domain $$\{x,y: x>y\}$$ That is, $$P(X>Y)=\lambda_1\lambda_2\int_0^{\infty}e^{-\lambda_1x}\int_0^xe^{-\lambda_2y}\ dy \ dx=$$ $$=\lambda_1\int_0^{\infty}e^{-\lambda_1x}(1-e^{-\lambda_2x})\ dx=1-\frac{\lambda_1}{\lambda_1+\lambda_2}.$$
If $g(x,y)=1$ if $x>y$ and $g(x,y)=0$ otherwise then:
$$P(X>Y)=\mathbb Eg(X,Y)=\int_0^{\infty}\int_0^{\infty}g(x,y)f_X(x)f_Y(y)dxdy=\int_0^{\infty}f_Y(y)\int_y^{\infty}f_X(x)dxdy=\int_0^{\infty}f_Y(y)P(X>y)dy$$
Work this out.