Question about exp. distribution

Question

We know that $X\sim \exp(1),Y\sim \exp(2)$ and they are independent.
What is $P(Y>X)$?
exp=Exponential...

Thank you!

Answer 1

From this previous question(pdf of the difference of two exponentially distributed random variables) they derive for $Y\sim Exponential(\lambda)$ and $X\sim Exponential(\mu)$ letting $Z=X-Y$ a CDF of $Z$ $$P(X<Y)=P(X-Y<0)=P(Z<0)=F_{Z}(0)=1-\frac{\lambda}{\lambda+\mu}e^{-\mu (0)}=1-\frac{\lambda}{\lambda+\mu}=\frac{\mu}{\lambda+\mu}$$

in your example $\lambda=2$ and $\mu=1$ so we have $P(Y>X)=\dfrac{1}{3}$

If a theoretical method seems out of your grasp, You can always simulate large sample and get rough estimate of probability (guaranteed to be close by Law of Large Numbers) here is some R code that could estimate it

X=rexp(10000,1)

Y=rexp(10000,2)

Z=Y-X

positiveZ=Z[Z>0]

prob=length(positiveZ)/length(Z)

prob

PS Thanks to Robert Isarel and Andre Nicolas for their solution to the CDF