Defining $P_B=P(A|B)$ provided $P(B)>0$. $P_B(A|C)$ is not necessarily well-defined for every event $C$ provided $P(C)>0$. Why not?

Question

I started learning probabilty about two weeks ago and I got this question:

Defining $P_B=P(A|B)$, provided $P(B)>0$.

$P_B(A|C)$ is not necessarily well-defined for every event $C$, provided $P(C)>0$. Why not?

I am not sure how I should approch the question because I dont even know how to translate $P_B(A|C)$, by definition $P(A|B)=\frac{P(A\cap B)}{P(B)}$ but how do I make use of the formula with 3 events.

Answer 1

You have defined $P_B(A):=P(A|B)=\frac{P(A\cap B)}{P(B)}$ for every event $A$. Note that $P_B$ is also a probability, that is,

• $P_B(\emptyset) = 0$,
• $P_B(A^c)=1-P_B(A)$ for every event $A$, and
• $P_B\left(\bigcup_iA_i\right) = \sum_iP_B(A_i)$ for every countable, disjoint collection of events $\{A_i\}$.

If you are not familiar with countable additivity, you can replace this last statement with

• $P_B(A\cup C) = P_B(A)+P_B(C)$ for all events $A,C$ with $A\cap C = \emptyset$.

This is not an equivalent statement, but it is sufficient for this level of probability.

Now, when is $P(A|B)$ defined? Whenever $P(B)>0$. So, when is $P_B(A|C)$ defined? We need $P_B(C)>0$ - which is not the same as $P(C)>0$! Specifically, since $P_B(C) = \frac{P(B\cap C)}{P(B)}$, we need $P(B\cap C)>0$ to define $P_B(A|C)$.

But what does $P_B(A|C)$ actually mean? Write out the definitions and simplify:

$$P_B(A|C) = \frac{P_B(A\cap C)}{P_B(C)} = \frac{\frac{P\big((A\cap C)\cap B\big)}{P(B)}}{\frac{P(B\cap C)}{P(B)}} = \frac{P\big(A\cap (B\cap C)\big)}{P(B\cap C)} = P(A|B\cap C).$$

That is, if you first condition on $B$, then condition on $C$, this is the same as conditioning on $B\cap C$, or "$B$ and $C$". Hopefully this makes intuitive sense!