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


  $$P\left(X_i=\min_kX_k\right)=\frac{\lambda_i}\lambda$$
If you have a sequence of independent and identically distributed continuous random variables, then the index of the smallest instance among them will be discretely uniformly distributed. &nbsp; Because any one of them may turn out to be the smallest among them with no bias.

$$\begin{align}\forall j\in\{2,..,n\}\,&amp;~\Bigl[\,\mathsf P(X_1{=}\min_{i=1}^n\{X_i\})~=~\mathsf P(X_j{=}\min_{i=1}^n\{X_i\})\,\Bigr]\\[2ex]\sum_{j=1}^n \mathsf P(X_j{=}\min_{i=1}^n\{X_i\})~&amp;=~1\\[2ex]\hline\therefore~\mathsf P(X_1{=}\min_{i=1}^n\{X_i\})~&amp;=~1/n\end{align}$$