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, 2020 at 7:56
  • 5
    $\begingroup$ 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, 2020 at 9:46
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    $\begingroup$ 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, 2020 at 14:22

1 Answer 1


$P(X_k=x)=0$ 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

  • 1
    $\begingroup$ 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, 2020 at 9:03
  • $\begingroup$ You actually condition on $X_k$, not on a single value. This is why you integrate. $\endgroup$
    – Sergio
    Sep 1, 2020 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, 2020 at 15:19
  • $\begingroup$ @bjb Note: Then the conditional probability has nothing to do with the event itself (Borel-Kolmogorov paradox), instead it depends intricately on the random variable. In very brief: Take the usual definition of $$\mathsf E(X\mid Y)$$ as $$\mathsf E(X\mid\sigma(Y))$$ where $\sigma(Y)$ is the sigma-algebra generated by $Y$. Then by the factorization lemma $$\mathsf E(X\mid Y)=f\circ Y$$ where $f$ is some measurable function that is uniquely given $Y_\#\mathsf P$-almost everywhere. We can then write (as abuse of notation) $$\mathsf E(X\mid Y=y)=f(y)$$ for $\mathsf Y_\#\mathsf P$-almost every $y$. $\endgroup$ Jan 1 at 10:46

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