0

Can someone tell me if I made any mistakes on this proof :

prove :$$E(X|A)=\frac{E[XI_a]}{P(A)}$$ where $I_a=1$ if event A occurs, else $I_a=0$.

$$E(XI_a)=I_aE(E(X|A))=I_a\sum_AE(X|A)p(A)=E(X|A)p(A)$$ With the last equality being true because $A\neq0=>I_a=0$ $$E(X|A)p(A)=E(XI_a)=>E(XI_a)/p(A)=E(X|A) $$

Frank
  • 880
  • 1
    What is your definition of $\mathbb E[X\mid A]$ ? To me $\frac{1}{\mathbb P(A)}\mathbb E[X\boldsymbol 1_A]$ is the definition. – Surb May 23 '18 at 08:38
  • The expectation of X given some event A occurs – Frank May 23 '18 at 08:59
  • I know what it mean ! But how do you defined it ? To me, the definition of $\mathbb E[X\mid A]$ is precisely $\frac{1}{\mathbb P(A)}\mathbb E[X\boldsymbol 1_A]$. By the way, Wikipedia has the same definition as me as you can see here – Surb May 23 '18 at 09:17
  • You can verify first for an indicator function, then to step function, and using the machinery to positive random variable and all random variables etc. – Gordon May 24 '18 at 20:31

0 Answers0