1
$\begingroup$

I want to prove, that: $$ E[ E[X |Y]]= E[X] $$ for X,Y discrete random variables.

Proof: $$ E[ E[X |Y]] = \sum_y E[X|Y=y] \cdot P(Y=y) = \sum_y \sum_x \frac{P(X=x, Y=y)}{P(Y=y)}\cdot P(Y=y)$$

$$ \underbrace{=}_{Fubini} \sum_x x \sum_y P(X=x, Y=y)=? $$

Why should be $\sum_y P(X=x, Y=y) = P(X=x)$ ?

$\endgroup$
6
  • 1
    $\begingroup$ It follows directly from definition, unless perhaps you work only on discrete random variables, and you defined the concept for them? $\endgroup$
    – Jakobian
    Commented Jul 17, 2019 at 14:06
  • $\begingroup$ Which defintion do you mean? $\endgroup$
    – Sarah
    Commented Jul 17, 2019 at 14:07
  • 1
    $\begingroup$ $E[X|Y]$ is an $Y$ measurable random variable such that for every $Y$-measurable set $A$ we have $E[1_AE[X|Y]] = E[1_AX]$. This is the definition. In particular we can take $A = \Omega$. $\endgroup$
    – Jakobian
    Commented Jul 17, 2019 at 14:09
  • $\begingroup$ Sorry, I do not know this definition. Is there any argument for $\sum_y P(X=x, Y=y) = P(X=x)$ $\endgroup$
    – Sarah
    Commented Jul 17, 2019 at 14:11
  • 1
    $\begingroup$ Yes, this is just countable additivity of measure $P$. $\bigcup_y\{X = x, Y = y\} = \{X = x\}$ $\endgroup$
    – Jakobian
    Commented Jul 17, 2019 at 14:12

1 Answer 1

2
$\begingroup$

$$P(X = x) = P(\cup_{y \in \Omega_y} \{X = x\}\cap \{Y=y\}) = \sum_{y \in \Omega_y}P(\{X =x\} \cap\{Y=y\}),$$ since $y$ can take only one value at a time and $\Omega_y$ is countable. I.e. $$(\{X =x\} \cap\{Y=k\}) \cap(\{X =x\} \cap\{Y=l\}) = \emptyset$$ if $k \neq l$.

$\endgroup$
3
  • $\begingroup$ You can use \{ and \} for brackets to show up. $\endgroup$
    – Jakobian
    Commented Jul 17, 2019 at 14:15
  • $\begingroup$ Thank you,I've wanted to find that out for a while. $\endgroup$ Commented Jul 17, 2019 at 14:16
  • $\begingroup$ @GradaGukovic: Thank you very much:) $\endgroup$
    – Sarah
    Commented Jul 17, 2019 at 14:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .