Prove $P(X\mid Y)= P(X\mid Y\cap E_i)P(E_i\mid Y)$

Question

Assume that $E_1,…,E_n$ are events that partition the sample space $S$. Let X and Y be events in $S$ such that $P(Y\cap E_i)>0$ $\forall 1\le i\le n$

I would like to show that

$$P(X\mid Y)= P(X\mid Y\cap E_i)P(E_i\mid Y)$$

How can I prove that?

Answer 1

Suppose that $X = E_1$, $Y = S$ and $E_2 = S \backslash E_1$. Hence $P(X\mid Y)= 1$, $P(X\mid Y\cap E_2) = P(E_1 \mid E_2) =0$ and $$P(X\mid Y)= 1 \ne 0 = P(X\mid Y\cap E_i)P(E_i\mid Y)$$ for $i=2$.

But $$P(X\mid Y)= \sum_{i=1}^n P(X\mid Y\cap E_i)P(E_i\mid Y)$$ as $$P(X\mid Y) = \frac{P(X \cap Y)}{P(Y)}= \frac{\sum_{i=1}^n P(X \cap Y\cap E_i)}{P(Y)}$$ $$= \sum_{i=1}^n \frac{ P(X \cap Y\cap E_i)}{P(Y\cap E_i)} \frac{P(Y\cap E_i)}{P(Y)} = \sum_{i=1}^n P(X\mid Y\cap E_i)P(E_i\mid Y)$$