Let $X$ and $Y$ be independent exponential random variables with rates $\mu_1$ and $\mu_2$

Question

Let $X$ and $Y$ be independent exponential random variables with rates $\mu_1$ and $\mu_2$.

a. What is the distribution of $\min(X,Y)$?

b. Find $P(X<Y)$

c. What is $P( X<Y | \min(X,Y)=z )$?

I've been struggling to work through this problem for a couple days. I think I have a part of it done: $$(u1+u2)e^{-u1+u2}z,$$ where $z = \min(x,y)$. But it would be great to get some help and explanation.

Answer 1

a. The easiest way is to first find the cdf, and then take the derivative to get the cdf. Let $Z=\min(X,Y)$. $$P(Z > z) = P(X>z \text{ and } Y>z) = P(X>z) P(Y>z) = e^{-\mu_1 z} e^{-\mu_2 z} = e^{-(\mu_1+\mu_2) z}.$$ See if you can finish from here. [Although you seem to already have gotten the answer.]

Then the density is $f_Z(z) = \frac{d}{dz} P(Z\le z) = \frac{d}{dz}(1-e^{-(\mu_1+\mu_2)z}) = (\mu_1+\mu_2)e^{-(\mu_1+\mu_2)z}$.

b. Do the appropriate double integral. $$P(X<Y) = \int_0^\infty \int_x^\infty \mu_1 e^{-\mu_1 x} \mu_2 e^{-\mu_2 y} \mathop{dy} \mathop{dx} = \cdots$$

$=\int_0^\infty \mu_1 e^{-\mu_1 x} e^{-\mu_2 x} \mathop{dx} = \frac{\mu_1}{\mu_1+\mu_2}$.

c. Informally, $$P(X<Y \mid Z=z) = \frac{P(X<Y, Z \in dz)}{P(Z \in dz)} = \frac{P(X \in dz, Y>z)}{P(Z \in dz)} = \frac{\mu_1 e^{-\mu_1 z} \mathop{dz} e^{-\mu_2 z}}{(\mu_1+\mu_2) e^{-(\mu_1+\mu_2) z} \mathop{dz}}=\frac{\mu_1}{\mu_1+\mu_2}.$$