# 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

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 $$Pr(X_k\in dt\ |\ X_k = X) = Pr(X_k = X)\times (*) = \lambda exp\left(-t\lambda\right)dt.$$