$X=\min\{X_1, X_1\cdot X_2, X_1\cdot X_2\cdot X_3, \ldots, X_1\cdot X_2 \cdots X_N\}$ where $X_i, \forall i\in\{1,2,...,N\}$ are continuous random variables, $X_i\geqslant 0$, e.g., exponential distribution.

I want to find the CDF of $X$ for any general $N$, i.e., $F_X(x)$.

For example, when $N=2, X=\min\{X_1, X_1\cdot X_2\}$, then we can write $$F_X(x) = \mathsf P(X < x) = \mathsf P(X_1 < x \mid X_2 \geqslant 1) + \mathsf P(X_1\cdot X_2 < x \mid X_2 < 1).$$

Can any one help me to write $F_X(x)$ for any $N$?

  • $\begingroup$ For a start $F_X(x) = \mathsf P(x_1<x \mid x_2\geq 1)\mathsf P(x_2\geq 1)+\mathsf P(x_1\cdot x_2 < x\mid x_2< 1)\mathsf P(x_2<1)$ $\endgroup$ – Graham Kemp Mar 31 '15 at 0:44
  • $\begingroup$ When N=2, it is OK. But I need the CDF valid for any N. $\endgroup$ – Frey Mar 31 '15 at 0:51

So for $N=3$ we have

$$\begin{align} F_X(x) & = \mathsf P(X_3\geq 1, X_2\geq 1)\mathsf P(X_1< x\mid X_3\geq 1, X_2\geq 1)\\ & + \mathsf P(X_3\geq 1, X_2< 1)\mathsf P(X_1\cdot X_2 < x\mid X_3\geq 1, X_2< 1) \\ & + \mathsf P(X_3< 1)\mathsf P(X_1\cdot X_2\cdot X_3< x\mid X_3< 1) \end{align}$$

So in general. $$\begin{align} F_X(x) & = \mathsf P(\bigcup_{j=2}^{N} \{X_j\geq 1\})\mathsf P(\{X_1 < x\}\mid \bigcup_{j=2}^{N} \{X_j\geq 1\}) \\ & + \sum_{k=2}^{N}\mathsf P(\{X_k < 1\}\cup \bigcup_{j=k+1}^{N}\{X_j \geq 1\})\mathsf P(\{\prod_{i=1}^k X_i < x\}\mid \{X_k < 1\}\cup \bigcup_{j=k+1}^{N}\{X_j \geq 1\}) \end{align}$$

| cite | improve this answer | |
  • $\begingroup$ Thanks Graham. Can you please check the first term which includes index "j" but I do not see it in the term ? $\endgroup$ – Frey Mar 31 '15 at 1:21
  • $\begingroup$ Thanks @Frey . The typo has been correxted. $\endgroup$ – Graham Kemp Mar 31 '15 at 1:48
  • $\begingroup$ This works Graham :) Great help !!! $\endgroup$ – Frey Mar 31 '15 at 1:56
  • $\begingroup$ Sorry to revive a thread from 5 years ago, but are we sure that this works? (I will give a -1 for now since I am highly uncertain) This seems like trying to partition the minimum into three cases; however, I don't think it would be as clean. (In particular, just because the last entry is less than 1, I don't see why the minimum should be $X_1 \cdot X_2 \cdot X_3$. It could be the case that $X_2$ is so large that the minimum is $X_1$. ) $\endgroup$ – E-A May 11 at 2:01
  • $\begingroup$ Hmmmmm......... $\endgroup$ – Graham Kemp May 11 at 20:54

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.