# Math Help - Iterated expectations

1. ## Iterated expectations

Let $I_A$ be the indicator function of the event A

$E[I_A]=E[E[I_A|A]]=E[E[1]]=1$

$E[I_A]=E[E[I_A|A^C]]=E[E[0]]=0$

:S

What's going on here?

2. Originally Posted by HD09
Let $I_A$ be the indicator function of the event A

$E[I_A]=E[E[I_A|A]]=E[E[1]]=1$

$E[I_A]=E[E[I_A|A^C]]=E[E[0]]=0$

:S

What's going on here?
With a complete system of events {A_1, A_2, ..., A_n}, we have formula:
$EX = \sum_{i=1}^n P(A_i) E(X | A_i)$
So I think
$E[I_A]=P(A) E[I_A|A]+ P(A^c) E[I_A|A^C]$
$=P(A) E[1|A]+ P(A^c) E[0|A^C]=p(A)$

3. I agree but I'm not really after THE answer so much as the reason why taking iterated expectations is giving me the wrong answer

4. Hello,
Originally Posted by HD09
Let $I_A$ be the indicator function of the event A

$E[I_A]=E[E[I_A|A]]=E[E[1]]=1$

$E[I_A]=E[E[I_A|A^C]]=E[E[0]]=0$

:S

What's going on here?
I guess the mistake is that in the iterated expectations, you have $\mathbb{E}[X]=\mathbb{E}[\mathbb{E}[X|Y]]$, where Y is a random variable, and not an event. And X has to be a random variable too.
Here, $I_A$ is indeed a random variable. But A is defined as an event.

A few facts :
X|A is a random variable. E(X|A) is a number.
X|Y is a random variable. E(X|Y) is a random variable. E(E(X|Y)) is a number.

5. Thanks! That's made it much clearer

6. $e(i_a)=p(a)$ won't stay upper case for some reason ....

$E(I_A)=P(A)$