In Wikipedia, for independent exponentially distributed random variables $X_1, \cdots ,X_n$ with rate parameters $\lambda_1, \cdots ,\lambda_n$, The probability $P(I=k)$ where $I=\textrm{argmin }_{i\in\{1,\cdots ,n\}}\{X_1,\cdots X_n\}$ were calculated as follows:

$\begin{align} P(I=k)& =\int_{0}^{\infty} P(X_k =x)P(X_{i\neq k}>x)dx \\ &=\int_{0}^{\infty}\lambda_k e^{-\lambda_k x}\left(\prod_{i=1,i\neq k}^{n}e^{-\lambda_i x}\right)dx \\ &= \lambda_k \int_{0}^{\infty}e^{-(\lambda_1+\cdots +\lambda_n )x}dx \\ &=\frac{\lambda_k}{\lambda_1+\cdots + \lambda_n}\end{align}$

However, I have a question about the first line. Isn't $P(X_k=x)=0$, as $X_k$ is a continuous random variable? How can we rigorously prove the first line and the second line?

  • 1
    $\begingroup$ Of course $P(X_k=x)=0$ for every $x$. The density $f_{X_k}(x)$ makes more sense instead of $P(X_k=x)$. $\endgroup$ Sep 1 '20 at 7:56
  • 4
    $\begingroup$ It looks like the Wikipedia page is using an abuse of notation, by using the generic function $\text{Pr}$ in a loose sense to refer either to a probability or a density, depending on the argument. In cases where the argument event is an equation (as opposed to an inequality), you should interpret it as a reference to the density. $\endgroup$
    – Ben
    Sep 1 '20 at 9:46
  • 1
    $\begingroup$ For some insight, you may think of this question in the following setting: you run a homogeneous Poisson process of rate $\lambda=\lambda_1+\cdots+\lambda_n$ and randomly label each event with the value $k$ with probability $p_k=\lambda_k/\lambda.$ This "thinning" yields $n$ independent Poisson processes with rates $p_k\lambda=\lambda_k.$ The question asks for the chance that the first event is labeled with $k.$ Obviously the answer is $p_k$! $\endgroup$
    – whuber
    Sep 1 '20 at 14:22

$P(X_k=x)=0$ for every $x$, but you can condition on $X_k=x$, $x \in [0,\infty)$:

\begin{align*} P(I=k) &= P(X_i>X_k, i\ne k) \\ &= \int_0^\infty P(X_i>X_k, i\ne k\mid X_k=x)\lambda_ke^{-\lambda_k x}dx\\ &= \int_0^\infty P(X_i>x, i\ne k)\lambda_ke^{-\lambda_k x}dx \\ &= \int_0^\infty \lambda_ke^{-\lambda_k x}dx\left(\prod_{i\ne k}e^{-\lambda_i x}\right)dx \\ &\text{etc.} \end{align*}

See https://mast.queensu.ca/~stat455/lecturenotes/set4.pdf

  • $\begingroup$ What does $P(X_k > X_k , i\neq k |X_k =x)$ mean here? using the standard definition of conditional probability, Isn't it $P(X_k > X_k , i\neq k , X_k=x)/P(X_k=x) = (something)/0 ?$ $\endgroup$
    – Kaira
    Sep 1 '20 at 9:03
  • $\begingroup$ You actually condition on $X_k$, not on a single value. This is why you integrate. $\endgroup$
    – Sergio
    Sep 1 '20 at 9:29
  • $\begingroup$ @Kaira Conditional probability on an event that has probability zero can make sense if you use a measure theoretic definition of conditional probability. $\endgroup$
    – bjb568
    Sep 1 '20 at 15:19

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.