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.