4
$\begingroup$

I have to multiply this:

$P(a|b,c)·P(b|c)$

How do you multiply those two expressions?

It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.

$\endgroup$
  • $\begingroup$ Intuitively you could compare this with P(a|b)P(b) = P(a,b) $\endgroup$ – Martijn Weterings Jan 8 at 20:48
  • $\begingroup$ Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?" $\endgroup$ – Alexis Jan 8 at 20:52
  • $\begingroup$ The nomenglature. $\endgroup$ – VansFannel Jan 8 at 20:55
  • $\begingroup$ Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$. $\endgroup$ – StatsStudent Jan 8 at 21:08
  • $\begingroup$ @VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers. $\endgroup$ – StatsStudent Jan 8 at 22:40
8
$\begingroup$

Those two expressions simply mean:

(the Probability of event $a$ given the events $b$ and $c$) $\times$ (the Probability of event $b$ given event $c$).

The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:

$P(a|b,c)·P(b|c)=0.40\times0.70=0.28$

UPDATED BASED ON YOUR EDITED QUESTION:

The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $\frac{P(DE)}{P(E)}$

\begin{eqnarray*} P(a|bc)P(b|c) & = & P(a|bc)\frac{P(bc)}{P(c)}\\ & = & \frac{P(abc)}{P(bc)}\frac{P(bc)}{P(c)}\\ & = & \frac{P(abc)}{P(c)}\\ & = & \frac{P[(ab)c]}{P(c)}\\ & = & P(ab|c) \end{eqnarray*}

So, in all, we have:

\begin{eqnarray*} P(a|bc)P(b|c) & = & P(ab|c) \end{eqnarray*}

$\endgroup$
3
$\begingroup$

If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule: $$ P(a,b,c) = P(a|b,c) P(b|c) P(c) $$ which would imply: $$ P(a|b,c) P(b|c) = \frac{P(a,b,c)}{P(c)} $$ so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.

$\endgroup$
  • $\begingroup$ I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities. $\endgroup$ – VansFannel Jan 8 at 20:45

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.