The Law of Iterated Expection Looks like E{E(X|Y)} when the partition you use is generated by the random variable Y rather than Ω. What happens when you use such a partition on the Law of Total probability?
1 Answer
$\begingroup$
$\endgroup$
3
Same thing: $$\Pr(A) = \operatorname E(\Pr(A\mid Y)). \tag 1$$
For each value $y$ that the random variable $Y$ can take, you have a conditional probability $\Pr(A\mid Y=y),$ and that is a function of the value $y,$ so call that $h(y).$ Then $h(Y)$ is a random variable an is what we call $\Pr(A\mid Y).$
Notice that $(1)$ is a consequence of the law of total expectation: It is just the case where $X$ is the indicator random variable of the event $A,$ i.e. $$ X = \begin{cases} 1 & \text{if } A \\ 0 & \text{if not } A. \end{cases} $$ Then you have $\operatorname E(X) = \Pr(A)$ and $\operatorname E(X\mid Y) = \Pr(A\mid Y).$
-
1$\begingroup$ Thank you! I got as far as P(A)= ∑P(A|Y=yi)pY(yi) over yi in curly Y but wasn't sure where to take it next. $\endgroup$– MattJan 18, 2018 at 16:49
-
$\begingroup$ @Matt : I'm glad it helped. You can "accept" the answer if it's appropriate. $\endgroup$ Jan 18, 2018 at 16:59
-
$\begingroup$ Oh yes, I'm still new to this site I'm afraid. $\endgroup$– MattJan 18, 2018 at 17:04