Distribution of the index of the variable which achieves the minimum of exponential random variables

I am reading Exponential distribution from Wiki, and it is said that the index of the variable which achieves the minimum is distributed according to the law $$P(k|X_k=min\{X_1,X_2,...,X_n\})=\frac{\lambda_k}{\lambda_1+...+\lambda_n}$$

I don't know how to prove this property. I try the case $n=2$ in different ways.

First, I find $P(X_1\le X_2)=\frac{\lambda_1}{\lambda_1+\lambda_2}$, but I can't change this to the conditional probability formally.

Second, I try to prove this through pdf. Let $Y=min\{X_1,X_2\}$, I want to calculate $f_{X_1|Y}(x_1,y)$. However I find that there should be infinite value of $f_{X_1|Y}(x_1,y)$ at $x_1=y$ , since the conditional probability is actually a discrete distribution. I don't know how to obtain the discrete distribution from continuous pdf.

Please tell me how can I continue my proof or give another formal proof, Thanks!!

it is said that the index of the variable which achieves the minimum is distributed according to the law $$P(k\mid X_k=\min\{X_1,X_2,...,X_n\})~=~\frac{\lambda_k}{\lambda_1+...+\lambda_n}$$

Y...eah.   That's a poor way to express it.   That is not actually a conditional probability.

All they are saying is if we let $$K$$ be the random variable defined as the index of the minimum value of the sample, that is $$X_K=\min\{X_1,X_2,\ldots,X_n\}$$, then the probability mass function of $$K$$ is : $$P(K=k) ~=~ \dfrac{\lambda_k}{\lambda_1+\lambda_2+\cdots+\lambda_n}~~\mathbf 1_{k\in\{1,2,\ldots,n\}}$$

So, for $$n=2$$ you have found $$P(X_1\leqslant X_2)~=~ P(K=1) ~=~ \dfrac{\lambda_1}{\lambda_1+\lambda_2}$$

In general \begin{align}P(K=k) ~&=~ \int_0^\infty f_{X_k}(t) \prod\limits_{j\in\{1..n\}\setminus\{k\}} (1-F_{X_j}(t))\operatorname d t \\[1ex] &\vdots\\[1ex] &=~ \dfrac{\lambda_k}{\sum_{j=1}^n \lambda_j}\end{align}

• Thank you, I still have a question. You said that it was not actually a conditional probability, but if I define another rv $Y=min\{X_1,X_2,...,X_n\}$, is $P(K=k|Y=y)$ a conditional probability? Can we say that $P(K=k|Y=y)=P(K=k), \forall y \ge 0$? – Jy Chen Mar 30 '17 at 11:23
• If $Y$ is the minimum value of the samples, $K$ is the index of the sample with the minimum value, then ... yes, it would appear that they actually are independent. @JyChen – Graham Kemp Mar 30 '17 at 13:01
• Yeah, I think so, thanks! – Jy Chen Mar 30 '17 at 16:38
• @GrahamKemp do you have a reference for the formula "In general"? – Daniel Korpi Kastinen Jun 7 '18 at 16:28
• @GrahamKemp A correction: it should be P(K=1) instead of P(K=2) as the index of the variable with miinimum value is 1. – Nikhil Chilwant Apr 21 at 10:41