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!!
 A: 
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 $$\begin{align}P(K=k) ~&=~ \int_0^\infty f_{X_k}(t) \prod\limits_{j\in\{1..n\}\setminus\{k\}} (1-F_{X_j}(t))\operatorname d t \\[1ex] &\vdots\\[1ex] &=~ \dfrac{\lambda_k}{\sum_{j=1}^n \lambda_j}\end{align}$$
A: The probability density of an exponential random variable with parameter $\lambda$ is $\lambda e^{-\lambda x}$ which is equal to $\lambda$ at $0$. It means that the probability that it will be less than $\delta$ is equal to $\lambda \delta$ for $\delta \rightarrow 0$. So if we have $n$ independent exponential random variables with parameters $\lambda_i$ the probability that any of them will be less than $\delta$ is equal to $\lambda_i\delta$ for $\delta \rightarrow 0$. Now, the probability that two of them are less than $\delta$ is significantly smaller than any of the $\lambda_i\delta$ for $\delta \rightarrow 0$, thus the probability that the $i$-th variable will be smaller than $\delta$ conditioned on the event that one of them is approaches:
$$\frac{\lambda_i\delta}{\lambda_1\delta+\lambda_2\delta+\dots+\lambda_n\delta}=\frac{\lambda_i}{\lambda_1+\lambda_2+\dots+\lambda_n}$$
as $\delta \rightarrow 0$. Therefore, by memoryless property, it is also the probability that the $i$-th variable will be the smallest one.
