Is there a convenient way to express $$ p(A\leq a |B\leq b \land C\leq c), $$ when all I've got is an expression for $$ p(A\leq a |B = b \land C = c), $$ $$ p(A\leq a |B = b), $$ and $$ p(A\leq a |C = c)? $$

I do have an expression for the probability mass functions of $A$, $B$ and $C$, and their cumulative distribution functions, but I would like to avoid evaluating them, if possible.


I should mention also that I am not able to use Bayes' theorem straight-away, because I am not able to compute $p(B\leq b \land C\leq c)$.


No, I'm afraid that's not possible. Consider the simpler problem of only two random variables. You want to evaluate $$ P(A\leq a|B\leq b) $$ but only have access to $P(A\leq a|B=b)$ which does not contain enough information to evaluate the expression you are after. However, \begin{align} P(A\leq a|B\leq b)&=\frac{P(B\leq b|A\leq a)P(A\leq a)}{P(B\leq b)}\\ &=\frac{P(A\leq a)}{P(B\leq b)}\int^b db'P(B=b'|A\leq a)\\ &=\frac{P(A\leq a)}{P(B\leq b)}\int^b db'\frac{P(A\leq a|B=b')P(B=b')}{P(A\leq a)}\\ &=\frac{1}{P(B\leq b)}\int^b db'P(A\leq a|B=b')P(B=b'). \end{align}

In other words, you need to evaluate the cumulative distribution function.

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  • $\begingroup$ Thank you! Yes, I see.. This is bad news for me, since it will take a long time to run. Maybe I can find other ways to speed up the computations. Just a follow up: Would this be correct in order to extend this to the three variable-case: $p(A\leq a|B\leq b \land C\leq c) = p(B\leq b\land C\leq c)^{-1}\int^c\int^b p(A\leq a|B=b'\land C=c')p(B=b'\land C=c') db' dc'$? $\endgroup$ – Tommy L Feb 26 '16 at 11:31
  • $\begingroup$ Yes, that looks good. Can you draw samples from your distributions? Doing the integral using Monte Carlo methods might be quicker than fiddling with numerical integrators. $\endgroup$ – Till Hoffmann Feb 26 '16 at 11:34
  • $\begingroup$ Perfect! Then I'll try it using this form. My variables are actually discrete, so I'll just sum over $b'$ and $c'$. There may be many levels for both $b'$ and $c'$, but I think it will be manageable. The main problem is the joint probability in the denominator. I would have to use the chain rule and compute it recursively, which I believe will take a long time. Some variables are independent, or nearly indepenent, though, which may help me somewhat in pruning the recursive tree. $\endgroup$ – Tommy L Feb 26 '16 at 11:45

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