Let $X$ and $Y$ be independent random variables of exponential distribution with mean $1$. Find the mean of $Z=\min \{X,Y\}$.

Question

Let $X$ and $Y$ be independent random variables of exponential distribution with mean $1$. Another variable $Z$ is defined as $\min \{X,Y\}$. I need to find the mean of $Z$.

I only figured that $Z=x,Z=y$ with equal probabilty $0.5$, not sure what to do now.

Thanks in advance.

It gave me my answer,thanks – Frogfire Sep 15 '19 at 19:28 — Frogfire, Sep 15 '19 at 19:28

score 0 · Answer 1 · answered Sep 15 '19 at 19:22

0

The distribution of the $\min[X, Y]$ (given your conditions) is an exponential distribution with mean $0.5$.

answered Sep 15 '19 at 19:22

David G. Stork

29,774

score 0 · Answer 2 · answered Sep 15 '19 at 19:32

0

$F_{min(X, Y)}(x) = \mathbb{P}(min(X, Y) < x) = P(X < x, Y < x) = F_{X}(x) \cdot F_{Y}(x) = \{ \text{identically distributed} \} = F_{X}(x)^2 = e^{-2x} \cdot \mathbb{I}[x \ge 0]$. Then you just need to compute the expectation of this distribution

answered Sep 15 '19 at 19:32

Joitandr

998

Let $X$ and $Y$ be independent random variables of exponential distribution with mean $1$. Find the mean of $Z=\min \{X,Y\}$.

2 Answers2