$A = \min(A_1, ..., A_n)$ where $A_i \sim Exp(\lambda_i)$ indep. find $P(A = A_k)$

Question

I have calculated that $A \sim Exp(\sum_{i=1}^n \lambda_i)$ but to find $P(A= A_k)$ is proving difficult as I am not sure how to calculate the distribution of $A-A_k$, perhaps there is a better way to do this?

Answer 1

Recall that $B_n\sim\mathsf{Exp}(\sum_{i=1}^n \lambda_i)$ (here I am writing $B_n:=\min_{1\leqslant i\leqslant n}A_i$). This can be seen easily by induction as clearly the statement is true for $n=1$, and assuming it to be true for some positive integer $n$, then for $t>0$ we have \begin{align} \mathbb P(B_{n+1}> t) &= \mathbb P(B_n> t, A_{n+1}> t)\\ &= \mathbb P(B_n> t)\mathbb P(A_{n+1}> t)\\ &= e^{-\sum_{i=1}^n \lambda_i}e^{-\lambda_{n+1}}\\ &= e^{-\sum_{i=1}^{n+1} \lambda_i}, \end{align} so that $B_{n+1}\sim\mathsf{Exp}(\sum_{i=1}^{n+1} \lambda_i)$.

Now, given $1\leqslant i\leqslant n$, we have $$ \mathbb P(A=A_i) = \frac{\lambda_i}{\sum_{j=1}^n \lambda_j}, $$ as if $A_i\sim\mathsf{Exp}(\lambda_i)$ and $B_i\sim\mathsf{Exp}(\lambda_j)$ are independent then $$ \mathbb P(A_i\leqslant A_j) = \int_0^\infty\int_0^t \lambda_i e^{-\lambda_i s}\lambda_j e^{-\lambda_j t}\ \mathsf ds\ \mathsf dt = \frac{\lambda_i}{\lambda_i+\lambda_j}. $$