Lets say I'm given a table of conditional probabilities. For example, lets say that we have probabilities regarding human age. p(3|2) is probability of an individual reaching age 3 given that she is age 2.

Only thing I have is a chain. p(4|3), p(3|2), -- p(n+1 | n).

What probabilities can I calculate given just these conditional probabilities?

  • $\begingroup$ Can we assume that p(2|3)=1? $\endgroup$
    – Dan
    Commented Dec 10, 2015 at 10:37

1 Answer 1


Using $P(A\cap B)={P(A)}{P(B|A)}$ you can get ratios of the form $\frac{P(3)}{P(2)},\frac{P(4)}{P(3)},$ etc.

Namely: Let $A=2$ and $B=3$, then

  • $P(A\cap B)=P(2\cap 3)=P(3)$, since reaching $2$ and $3$ and reaching $3$ are the same events.
  • This gives $P(3)={P(2)}{P(3|2)}$, so $\frac{P(3)}{P(2)}=P(3|2)$ and $P(3|2)$ you know.

Other than that, I think you cannot derive anything meaningful. Ofcourse, if you know some chance $P(n)$ explicitly, you can calculate every other $P(m)$, $m\in\mathbb{N}$.

  • 2
    $\begingroup$ $\mathsf P(A\cap B) = \mathsf P(A)\cdot\mathsf P(B\mid A)$ $\endgroup$ Commented Dec 10, 2015 at 11:09
  • $\begingroup$ oh crap! Thank,s gonna edit it. $\endgroup$
    – Eric S.
    Commented Dec 10, 2015 at 11:11

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