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$X=\min\{X_1, X_1\cdot X_2, X_1\cdot X_2\cdot X_3, \ldots, X_1\cdot X_2 \cdots X_N\}$ 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 $N$?

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

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  • $\begingroup$ 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

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