Independence Exponential Distribution

Question

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 ?

Answer 1

$$P(X>Y)=\int_0^{\infty}P(Y<X\mid X=x)\lambda_1e^{-\lambda_1 x}\ dx=$$

$$=\int_0^{\infty}P(Y<x)\lambda_1e^{-\lambda_1 x}\ dx=\int_0^{\infty}(1-e^{-\lambda_2 x})\lambda_1e^{-\lambda_1 x}\ dx=$$ $$=\lambda_1\int_0^{\infty}e^{-\lambda_1 x}\ dx-\lambda_1\int_0^{\infty}e^{-x(\lambda_1+\lambda_2)}\ dx=$$ $$=1-\frac{\lambda_1}{\lambda_1+\lambda_2}.$$

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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,

$$P(X>Y)=\lambda_1\lambda_2\int_0^{\infty}e^{-\lambda_1x}\int_0^xe^{-\lambda_2y}\ dy \ dx=$$

$$=\lambda_1\int_0^{\infty}e^{-\lambda_1x}(1-e^{-\lambda_2x})\ dx=1-\frac{\lambda_1}{\lambda_1+\lambda_2}.$$

Answer 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.