[Math] Probability exponential random variable smallest among others.

Question

For $X_1,\dots,X_n$ exponential random variables with mean $E(X_i)=\mu_i$. Now I want to calculate the probability that $X_i$ is the smallest among $X_1,\dots,X_n$. Therefore I am trying to calculate the $P(X_i=\min({X_1,\dots,X_n}))$.
Now I have already calculated the distribution of the random variable $Y_n=\min({X_1,\dots,X_n})$ which can be expressed by $P(Y_n\leq t)=1-(e^{-\mu t})^n$.
I assume I need to use this with conditional probability in some way, but I do not exactly know how. Anyone has suggestions? thx

Answer 1

Fix some $i$ and note that, for every $x$ and every $k$, $$P(X_k>x)=e^{-\lambda_kx}$$ where $$\lambda_k=\frac1{\mu_k}$$ hence $$P\left(\min_{k\ne i}X_k>x\right)=P\left(\bigcap_{k\ne i}\{X_k>x\}\right)=\prod_{k\ne i}e^{-\lambda_kx}=e^{-(\lambda-\lambda_i)x}$$ where $$\lambda=\sum_i\lambda_i$$ Since $X_i$ is independent of $(X_k)_{k\ne i}$, this implies $$P\left(\min_{k\ne i}X_k>X_i\,{\large\mid}\, X_i\right)=e^{-(\lambda-\lambda_i)X_i}$$ Integrating both sides with respect to the distribution of $X_i$, one gets $$P\left(\min_{k\ne i}X_k>X_i\right)=E\left(e^{-(\lambda-\lambda_i)X_i}\right)=\int_0^\infty e^{-(\lambda-\lambda_i)x}\lambda_ie^{-\lambda_ix}dx$$ and finally, for every $i$,

$$P\left(X_i=\min_kX_k\right)=\frac{\lambda_i}\lambda$$