2
$\begingroup$

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)?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
1
  • 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
3
$\begingroup$

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 $(1).$

$\endgroup$
4
  • $\begingroup$ What does & mean in (1) and is there a name for this formula to learn more? $\endgroup$ Aug 19 '17 at 17:42
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

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.