2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 next step depends on the level of sophistication you are used to and, since you say nothing about this, it is difficult to continue but...

...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, $$p_i=\int_0^\infty f_i(x)\exp\left(-\sum_{j\ne i}\lambda_j x\right)\mathrm dx=\int_0^\infty \lambda_i\exp\left(-\lambda x\right)\mathrm dx=\frac{\lambda_i}{\lambda}.$$

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Did Sep 2 '14 at 19:19
  • $\begingroup$ @user You quite recently posted a question, whose formulation suffers from the same basic flaws as your answer here. An answerer over there made this fact quite clear, let me suggest to meditate their answer. $\endgroup$ – Did Sep 2 '14 at 19:22
  • $\begingroup$ Yes, I agree. Please take a look at my attempt to fix the above answer. $\endgroup$ – vdesai Sep 2 '14 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.