Law of total probability with extra conditioning, question regarding assumptions

Question

I'm trying to understand the highlighted condition in the theorem above. And I'm having a difficult time understanding why it's there. The theorem is discussing law of total probability with extra conditioning. Now in regular law of total probability, all we require is that our set of $A{_i}$ must partition the space, I believe there's no requirement that: $all$ $A_i$ $>$$0$. So in this instance why must be it that be that $A_i$ $>$ $0$ for all $A_i$ in $E$?

Let's say for the sake of argument there's some $A_i$ term that's not in $E$. Given that we are conditioning on $E$, our term $P(A_i|E)$ for that $i$ will simply be zero right?

So in our summation we will simply multiply by $0$ and for that $i$ term the result will simply be zero and it won't matter? And yet the condition is listed so I feel it must be there for a good reason, but I don't understand why.

Answer 1

The requirement that $\Pr[A_i \cap E] > 0$ is because if it is equal to zero, the event $A_i \cap E$ never occurs, hence $$\Pr[B \mid A_i, E] = \frac{\Pr[B \cap (A_i \cap E)]}{\Pr[A_i \cap E]}$$ is undefined.

That said, it is important to note that the product

$$\Pr[B \mid A_i, E] \Pr[A_i \mid E] = \frac{\Pr[B \cap (A_i \cap E)]}{\Pr[A_i \cap E]} \cdot \frac{\Pr[A_i \cap E]}{\Pr[E]} = \frac{\Pr[B \cap A_i \cap E]}{\Pr[E]}$$

is well-defined even if $\Pr[A_i] = 0$, so it is not a requirement to have a nontrivial partition of $S$ if we write

$$\Pr[B \mid E] = \frac{1}{\Pr[E]} \sum_{i=1}^n \Pr[B \cap A_i \cap E].$$

The only requirement for this formula to hold is that $\Pr[E] \ne 0$. Those terms for which $\Pr[A_i] = 0$ will simply result in those terms in the sum being ignored, as it should. So why don't we write the formula in this second form? The reason is the same reason why we write the (usual) law of total probability the way we do: because the conditional probabilities $\Pr[B \mid A_i]$ (or in this case, $\Pr[B \mid A_i, E]$) are known, for an appropriately selected partition of $S$. But doing so requires that we are not conditioning on events with zero probability.