# Law of total probability with extra conditioning, question regarding assumptions

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.

• $P(B\mid A_{i},\ E) = \frac{P(B\cap A_{i}\cap E)}{P(A_{i}\cap E)}$. Since $P(A_{i}\cap E)\in[0,1]$, what happens when $P(A_{i}\cap E)=0$ for at least one $i$? Commented Dec 23, 2023 at 23:43
• Dang. I completely ignored the first term in the summation, div by zero error. I see. Hmm, that's a bummer. Partitioning the space using Law of Total Prob in this way was a very handy trick, but I suppose I have to be more careful about it. Commented Dec 24, 2023 at 0:04
• Alternative remedy: $$P(B ~| ~E) = \sum_{A_i ~: ~A_i \cap E \neq \emptyset} P(B ~| ~A_i,E).$$ Commented Dec 24, 2023 at 0:50
• The formula is valid in all cases, even if $E=\emptyset$. It is enough to bear in mind that the conditional probability is defined except in a set of null measure. Commented Dec 24, 2023 at 6:50
• @Speltzu I don't quite follow. The formula is not defined if if there's an $A_i$ in $E$ that's 0 right? Commented Dec 24, 2023 at 16:32

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.

• Thank you very much. Still chewing your answer over a bit. I see the divide by zero issue, but then as you point out, it would cancel out. It seems like you're saying we don't need to have it as a requirement if we tweak the formula and remove the conditioning? Is the second part of your answer emphasizing the difference between the intersection of events vs their conditional relationship to each other? And that the the whole point of using LOTP is in cases where we know the conditional probability? I'm perhaps also not fully understanding the significance of your last sentence. Commented Dec 24, 2023 at 0:12
• With regards to your last sentence, why doesn't the same logic apply to regular LOTP. I think maybe I'm getting confused there as well. Commented Dec 24, 2023 at 0:15