# Minimum of exponential distributions

Consider $n$ independent random variables $X_i \sim \exp(\lambda_i)$ for $i = 1,\dots,n$. Let $\lambda = \sum_{i=1}^n \lambda_i$. Of course, the minimum of these exponential distributions has distribution:

$$X = \min_i \{X_i\} \sim \exp(\lambda),$$

and $X_i$ is the minimum variable with probability $\lambda_i/\lambda$. However, suppose I am given the fact that $X_a$ is the minimum random variable for some $a \in \{1,\dots,n\}$, so $X = X_a$. Knowing that, now what is the distribution of $X$? I suspect it would no longer be the case that $X \sim \exp(\lambda)$, but I am at a loss as to how to precisely figure out the distribution.

• Unless I'm missing something, doesn't Baye's rule for densities answer your question? Writing it out I get that $X$ conditioned on $X=X_a$ is indeed exponential with parameter lambda. – Jeremy Voltz Mar 25 '13 at 20:04
• Hardly MO stuff. Voting to close. – Did Mar 25 '13 at 20:30

## 2 Answers

The following link on the stats.SE answers your question in detail.

You might also find this wikipedia link useful.

As Jeremy Voltz suggested, you can solve this with the definition of conditional probability, and surprisingly, it does come out to be the same $Exp(\lambda)$ distribution.

$$Pr(X_k < s;\ \forall j\not= k,\ X_j > t) = (1 - exp(-s\lambda_k))exp\left(-t\sum_{j\not= k}\lambda_j\right).\\ Pr(X_k \in dt;\ \forall j\not= k,\ X_j > t) = -(-\lambda_k)exp(-t\lambda_k)exp\left(-t\sum_{j\not= k}\lambda_j\right)dt\\ = \lambda_kexp\left(-t\lambda\right)dt.\;\;\;(*)$$ So $$Pr(X_k\in dt\ |\ X_k = X) = Pr(X_k = X)\times (*) = \lambda exp\left(-t\lambda\right)dt.$$