Let $A_1, A_2, \ldots$ be independent random variables with known distributions.

I'm wondering if it's possible to equivalently decompose the probability above into separated constraints?
For example:
\begin{align*} \Pr(A_1 < b_1) & > 1-p_1 \\ \Pr(A_2 < b_2) & > 1-p_2 \\ &... \end{align*}


The answer is, in general, no.

Consider only two random variables, $A_1$ and $A_2$.

The constraint $A_1+A_2<c$ is the area below/to the left of a 45-degree line in the $(A_1,A_2)$-plane.

The constraints $A_1<b_1$ and $A_2<b_2$ are areas bounded by lines parallel to each of the axes. The events simply don't correspond.

In some particular circumstances (including known distributions for the $A$ variables, at the least) you may be able to derive some relationships, but there's no simple general relationship like you hope for.

  • $\begingroup$ Thank you. I met a optimization problem and the constraint is $Pr\left( \sum_{i}^{}{x_{i}+a_{i}}\leq c \right)\geq 1-p$ where $a_{i}$ is random variable. So I wanted to decompose the constraint into separated constraints: $Pr\left( x_{i}+a_{i}\leq b_{i} \right)\geq 1-p_{i}\; \forall i$ $\sum_{i}^{}{b_{i}}\leq c\; $ $ \sum_{i}^{}{p_{i}}\leq p$ It seems that they are not equivalent. Thank you. $\endgroup$ – fisher May 14 '14 at 5:52

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