# conditional probability exponential distributed

Let $X_1,\ldots,X_n$ be independent random variables, $X_i \sim{}$ $\mathrm{exponential}(\lambda_i)$. Let $X=\min\limits_{1\le i \le n} X_i$. Calculate $\mathbb{P}(X=X_i)$

At first I determined that $X\sim\mathrm{exponential}(\lambda_1+\cdots+\lambda_n)$. My next idea was that

$$\mathbb{P}(X=X_i\mid X\le x)=\frac{\mathbb{P}(X=X_i, X\le x)}{\mathbb{P}(X\le x)}=\frac{\mathbb{P}(X_i\le x)}{\mathbb{P}(X\le x)}=\frac{1-e^{\lambda_ix}}{1-e^{(\lambda_1+\cdots+\lambda_n)x}}$$

But I looked up in the solution and it should be $\mathbb{P}(X=X_i)=\dfrac{\lambda_i}{\lambda_1+\cdots+\lambda_n}$ a.s. Can someone help? Thanks, Zitrone

One can first clear up the situation, noting that, since the random variables $X_i$ are independent and absolutely continuous, there is almost surely no ex aequo, that is, $P(\exists i\ne j,X_i=X_j)=0$. Thus, the event $[X=X_i]$ is, up to some negligible events, equal to to the event $[\forall j\ne i,X_j\gt X_i]$.
...The most basic approach might be to consider the density $f_j$ of the distribution of each $X_j$ and to note that, by definition, $$p_i=P[\forall j\ne i,X_j\gt X_i]=\int_0^\infty f_i(x)\left(\prod_{j\ne i}\int_x^\infty f_j(x_j)\mathrm dx_j\right)\mathrm dx.$$ The value of the $j$th inner integral is $e^{-\lambda_jx}$ hence, introducing the parameter $\lambda=\lambda_1+\cdots+\lambda_n$, one gets, as desired, $$p_i=\int_0^\infty f_i(x)\exp\left(-\sum_{j\ne i}\lambda_j x\right)\mathrm dx=\int_0^\infty \lambda_i\exp\left(-\lambda x\right)\mathrm dx=\frac{\lambda_i}{\lambda}.$$
We know the following about exponentials: PDF: $f_{X_i}(x) = \lambda_i e^{-\lambda_i x}$ and $P(X_i > x) = e^{-\lambda_i x}$. We will make use of them below. Consider \begin{split} P(X_i = X) &= \int_{x=0}^{\infty} P(X_i = X, X \in (x-dx,x)) \\ &= \int_{x=0}^{\infty} P(X_i \in (x-dx, x), X_j > x\ \forall\ j\ne i) \\ &= \int_{x=0}^{\infty} P(X_i \in (x-dx, x)) \Pi_{j\ne i} P(X_j > x) \\ &= \int_{x=0}^{\infty} \lambda_i e^{-\lambda_i x} dx\ \Pi_{j\ne i} e^{-\lambda_j x} \\ &= \lambda_i \int_{x=0}^{\infty} e^{- x \sum_{j=1}^n \lambda_j}dx \\ &= \frac{\lambda_i}{\sum_{j=1}^n \lambda_j} \\ \end{split}
• It might be of interest to note that $$P(X_i = X, X = x)=P(X_i=x, X_j > x\ \forall\ j\ne i)=P(X_i=x)=0,$$ for every $i$ and every $x$. This suggests to try to devise an alternative, valid, approach of the question. – Did Sep 2 '14 at 19:19