Sum of conditional probabilities equals 1?

Question

Assuming that sum of probabilities for all possible events that can occur should sum to 1, how does one denote this for a conditional probability? Is it $P(A|E_1) + P(A|E_2) + ... = 1$, where $E_i$ is a specific event to be conditioned on? Or is the answer something else entirely?

Answer 1

This is a good example where intuition on conditional probability can be a good check of the formula. Suppose $P(A)=0$ (which is clearly possible); then we know intuitively that $P(A|E_i)=0$ for any $i$ (because event $A$ never happens, so how can it happen conditioned on something?), so the sum you gave will be $0$, not $1$.

But if the $E_i$ are a partition of the sample space, we do have the formula $$ P(E_1|A) + P(E_2|A) + \dots + P(E_n|A) = 1$$ with an intuitive explanation. Given that $A$ occurred, it's still true that one of the $E_i$ must occur, so the total probability of the $E_i$ occurring given $A$ must still be $1$.

Answer 2

If $E_1, E_2, \ldots$ is a collection of disjoint events whose union equals the entire sample space (exhaustive), then $P(A\cap E_1)+P(A\cap E_2)+\ldots = P(A|E_1)P(E_1)+P(A|E_2)P(E_2)+\ldots = P(A)$.

Answer 3

The previous answers are more than enough to understand what is going on.

I just want to state the general proposition (implicit in the answers) with a formal proof.

Proposition: Let $E_1,...,E_n,...$ be a countable collection of sets such that $E_i\cap E_j=\emptyset$ for all $i\neq j$ whose total union $\bigcup_{i=1}^\infty E_i$ is the whole sample space. If $p(A)>0$, then $$\sum_{i=1}^\infty p(E_i|A)=1.$$ *Remark: If $E_i\neq\emptyset$ for finitely many $i$'s, then the sum has finitely many positive terms.

Proof. Since $\{E_i\}_i$ is a countable collection of pairwise disjoint sets, then so is $\{E_i\cap A\}_i$. By the axiom of countable additivity, $$\sum_{i=1}^\infty p(E_i\cap A)=p\left(\bigcup_{i=1}^\infty (E_i\cap A)\right)=p(A).$$

By definition of conditional probability, $p(E_i\cap A)=p(E_i|A)\cdot p(A)$ for each $i$. Hence \begin{align*} \sum_{i=1}^\infty p(E_i|A)&=\sum_{i=1}^\infty \frac{p(E_i\cap A)}{p(A)}\\ &=\frac{1}{p(A)}\sum_{i=1}^\infty p(E_i\cap A)\\ &=1.\,_\blacksquare \end{align*}

Answer 4

One extra proof to complement the rest (the same but with a slightly different notation):

$$\sum_{i=0}^n P(E_{i}|A) = \sum_{i=0}^n(\frac{P(A|E_{i})P(E_{i})}{P(A)}) = \frac{1}{P(A)}\sum_{i=0}^n(P(A|E_{i})P(E_{i})) = \frac{P(A)}{P(A)} = 1 $$

The third step holds thanks to the product rule in probability.