Is conditional probability "transitive" (or how else this is called - please, explicate) in the meaning that $P(a \mid c)=P(a \mid b)P(b \mid c)$ ? Intuitively this seems so, but could you comment/proove on how to understand this (or refute)?


From the definition of conditional probability, we have $$ P(A\mid B)P(B\mid C) = \frac{P(A\cap B)}{P(B)} \frac{P(B\cap C)}{P(C)} $$ and $$ P(A\mid C) = \frac{P(A\cap C)}{P(C)}. $$ It's not clear to me why these should be equal. Let's find a counterexample.

Suppose we roll a fair die. Let $A$ be the event that $1$ shows, let $B$ be the event that an even number appears, and let $C$ be the event that either $1$, $2$, or $3$ shows. Then $$ P(A\mid C) = 1/3 \quad\text{and}\quad P(A\mid B) = 0 \quad\text{and}\quad P(B\mid C) = 1/3. $$ This example shows that the desired formula does not hold in general.

  • 2
    $\begingroup$ In the case where $A\subseteq B\subseteq C,$ however, the formulas on the right hand sides of the two equations both come out to $P(A)/P(C).$ So this formulation not only helps us find a counterexample in the general case, it shows the two probabilities are equal in the special case $A\subseteq B\subseteq C.$ It's a "two for the price of one" kind of answer! $\endgroup$
    – David K
    Aug 19 '17 at 20:35

Something of which that proposition is reminiscent is true: $$ \Pr(A\ \&\ B \mid C) = \Pr(A\mid B\ \&\ C)\Pr(B\mid C). \tag 1 $$

Now suppose $A\subseteq B\subseteq C.$ In that case we have $$ \Pr(A \mid C) = \Pr(A\mid B)\Pr(B\mid C) \tag 2 $$ since in the case $A\subseteq B\subseteq C,$ the proposition $(2)$ follows from $(1).$

  • $\begingroup$ What does & mean in (1) and is there a name for this formula to learn more? $\endgroup$ Aug 19 '17 at 17:42
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    $\begingroup$ It just means "and". The probability that $A$ and $B$ are both true, or, if you like, the probability that events $A$ and $B$ both occur, is $\Pr(A\ \&\ B)$ or $\Pr(A\cap B). \qquad$ $\endgroup$ Aug 19 '17 at 18:55
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    $\begingroup$ I don't know of a name for this identity, but it's not really different from saying $\Pr(A\ \&\ B) = \Pr(A\mid B) \Pr(B).$ All probabilities are conditional, and in this case we're conditioning on $C$ throughout. $\endgroup$ Aug 19 '17 at 18:56
  • $\begingroup$ That's a really helpful answer, thanks a lot! My software design was about to somewhat collapse because of this "intransitivity in general", but then you pointed the case when it's transitive, and fortunately it seems my case. $\endgroup$ Aug 19 '17 at 19:01

It is not so.

Maybe your intuition is that since with $A$, $B$, and $C$ being numbers we have:

$$\frac{A}{C}=\frac{A}{B} \cdot \frac{B}{C}$$

you may think something similar works for events as well.... but that is not true.

In fact, we don't even have:

$$P(A|B) \cdot P(B) = P(A)$$

for what we do have is:

$$P(A|B) \cdot P(B) = P(A \cap B)$$

So, if you are thinkng of treating the '$|$' as a kind of division ... don't!

In fact, $|$ is not an operator, and $A|B$ is not an event. It is a common mistake to see them that way, and I suspect you are doing something similar.


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