I am trying to find the pmf of a random variable of arg min of a sequence of independent exponential random variables i.e., $X_i = exp(\lambda_i)$.
Let L be a random variable defined by $$L = {arg min}_{1≤i≤n}X_i$$
That is, L = k if ${X}_{k}$ happens to be the minimum among $X_1,X_2,...,X_n$. I am interested in finding the pmf of L.
I am arriving at $\frac{\lambda_k} {\sum_{i=1}^n{\lambda_i}}$ as the PMF, where $\lambda_i$ is the exponential distribution parameter of $X_i$. I am trying to validate my answer here.
I assume $\lambda_i$ is a rate parameter, i.e. $$\Pr[X_i \le x] = 1 - e^{-\lambda_i x}, \quad x \ge 0.$$ Let $\Lambda = \sum_{i=1}^n \lambda_i$ be the sum of rates, and for the sake of convenience, let $\Lambda_k = -\lambda_k + \Lambda$; i.e., $\Lambda_k$ is the sum of all rates except the $k^{\rm th}$ rate. Then we observe that $$\prod_{i = 1 \\ i \ne k}^n \Pr[X_i \ge x] = e^{-\Lambda_k x}.$$
Because the $X_i$ are independent, $$\begin{align} \Pr[L = k] &= \int_{x=0}^\infty \Pr[L = k \mid X_k = x]f_{X_k}(x) \, dx \\ &= \int_{x=0}^\infty \left(\prod_{i \ne k} \Pr[X_i \ge x]\right)f_{X_k}(x) \, dx \\ &= \int_{x=0}^\infty e^{-\Lambda_k x} \lambda_k e^{-\lambda_k x} \, dx \\ &= \lambda_k \int_{x=0}^\infty e^{-\Lambda x} \, dx \\ &= \frac{\lambda_k}{\Lambda} \int_{x=0}^\infty \Lambda e^{-\Lambda x} \, dx \\ &= \frac{\lambda_k}{\Lambda}, \end{align}$$ which completes the proof.