Let $Y : (\Omega , \mathcal A,\Bbb P) \to \Bbb R$ a discrete variable and $A\in\mathcal A$, $P(A)\neq 0$, how can you prove only using elementary definitions the following: $$\frac{\Bbb E(Y1_A)}{\Bbb E(1_A)}=\sum_y y\frac{\Bbb P(A, Y=y)}{P(A)}$$

What I know is : $$E(1_A)=P(A)$$ and $$\Bbb E(Y) = \int_\Omega {Y(\omega) d\Bbb P (\omega)}=\int_\Bbb R{yd\Bbb P_Y(y)}=\sum_yy\Bbb P(Y=y)$$

(I am aware the denominators in both sides simplify but I leave it like that, as this is a formula used in the context of an introduction to conditional expectation.)

  • $\begingroup$ Do you have more information about $\mathcal{A}$? $\endgroup$ – Zhanxiong Nov 9 '15 at 5:08
  • $\begingroup$ No there isn't anything specific to $\mathcal A$. Should it matter for this problem? $\endgroup$ – KiwiKiwi Nov 9 '15 at 5:13
  • $\begingroup$ Oh, it seems needless. d.k.o. gave a nice answer. $\endgroup$ – Zhanxiong Nov 9 '15 at 5:14

By the Law of Total Expectation

$$\Bbb E(Y1_A)=\sum_{y}\Bbb E(Y1_A\mid Y=y)P\{Y=y\}$$ $$=\sum_{y}\Bbb yP\{A\mid Y=y\}P\{Y=y\}=\sum_{y}\Bbb yP\{A, Y=y\}$$


$$\Bbb E(Y1_A)=\sum_{y}\Bbb E(Y1_{A\cap\{Y=y\}})=\sum_{y}\Bbb yP\{A, Y=y\}$$

  • $\begingroup$ I am not sure how you do this : $$\sum_{y}\Bbb E(Y1_A\mid Y=y)P\{Y=y\}=\sum_{y}\Bbb yP\{A\mid Y=y\}P\{Y=y\}$$ $\endgroup$ – KiwiKiwi Nov 9 '15 at 5:23
  • $\begingroup$ @KiwiKiwi On $\{Y=y\}$ the value of $Y$ is known s.t. $$E(Y1_A\mid Y=y)=E(y1_A\mid Y=y)=yE(1_A\mid Y=y)=yP\{A\mid Y=y\}$$ $\endgroup$ – d.k.o. Nov 9 '15 at 5:26
  • $\begingroup$ Alright! Thanks! $\endgroup$ – KiwiKiwi Nov 9 '15 at 5:29

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