3
$\begingroup$

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.

Update:

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

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

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.