For each discrete random variable $X$ and a measurable set $B$, for which $P [B]> 0$, show that $E [X | B] = \frac{E [1_{B}X]} {P [B]}$. I have $E [X | B] = \frac{\sum x_{i} * P(\{X=x_{i}\} \cap B) } {P [B]}$ and I don't know what next.


$\frac{E[X 1_B]}{p[B]}=\frac{\sum_{\omega \in \Omega} x \times 1_B\times P(\{X=x\} ) } {P [B]}=\frac{\sum_{\omega \in B\cup B^{c}} x \times 1_B\times P(\{X=x\} ) } {P [B]}=\frac{\sum_{\omega \in B} x \times 1_B\times P(\{X=x\} )+0 } {P [B]}=\frac{\sum_{\omega \in B} x \times P(\{X=x\} ) } {P [B]}=\frac{\sum_{\omega \in B} x \times P(\{X=x\} \cap \{ \omega \in B\} ) } {P [B]}=\frac{\sum_{\omega \in \Omega} x \times P(\{X=x\} \cap \{ \omega \in B\} ) } {P [B]}=\sum_{\omega \in \Omega} x \times \frac{P(\{X=x\} \cap \{ \omega \in B\}}{P [B]} )=\sum_{\omega \in \Omega} x \times P(\{X=x\}|B )=E(X|B)$

another way

$\frac{E[X 1_B]}{p[B]}=\frac{EE[X 1_B|X]}{p[B]}$


$=\frac{\sum_x x g(x) p(X=x)}{p[B]}=\frac{\sum_x x E(1_B|X=x) p(X=x)}{p[B]}$

$=\frac{\sum_x x p(B|X=x) p(X=x)}{p[B]}$

$=\frac{\sum_x x p(X=x|B) p(B)}{p[B]}$

$=\frac{p(B) \sum_x x p(X=x|B) }{p[B]}$

$=\sum_x x p(X=x|B)=E(X|B)$

note that $E[1_B|X]$ is a random variable that it is a function of $X$.

  • $\begingroup$ What did you prove? $\endgroup$ – Kingis Mar 30 at 17:36
  • $\begingroup$ I edited it. is it what you want or not? $\endgroup$ – masoud Mar 30 at 19:15
  • $\begingroup$ Why do you use $\omega$? $\endgroup$ – Kingis Mar 30 at 19:28
  • $\begingroup$ since $E(X 1_B)=E(X 1_B(\omega))$ , $B$ is a measurable set so $B \in F$ in the probability space $(\Omega, F ,P)$ (so $B\subset \Omega$ such that $B\in F$ $\endgroup$ – masoud Mar 30 at 19:32
  • $\begingroup$ okey but i must write probably $\omega$ im my equation when i have X $\endgroup$ – Kingis Mar 30 at 19:47

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