Question

For an event A, let IA equal 1 if A occurs and let it equal 0 if A does not occur. For a random variable X, show that E[X|A] = E[XIA] P(A

Short answer

With the help of Lebesgue induction, we can prove this

Step-by-step solution

Given Information

let IA equal 1 if A occurs and let it equal 0 if A does not occur. For a random variable X

We have to show that E[X|A] = E[XIA] P(A

Explanation  using Lebesgue induction  

From the question, we will use Lebesgue induction over a random variable$X$, so

far the beginning let's assume that$X=I_M$for

some known set $B$, so we have:

$E(X\mid A)=E_a\left(I_a\right)$

$=P_A\left(I_g=1\right)$

$=P_A\left(B\right)$

$=\frac{P_A(A,B)}{P\left(A\right)}$

$=\frac{E\left(I_{\theta }{I}_A\right)}{P\left(A\right)}$

The claim for such a random variable holds, now suppose that$X$is the finite linear combination of indicator random variable:

$X=\sum _{j=1}^n{\alpha }_j{I}_n$

Step 3

Because of the linearity of conditional expectation and the first step of this induction, we have:

$E(X\mid A)=E\left(\sum _{j=1}^n {\alpha }_j{I}_{B_j}\mid A\right)$

$=\left(\sum _{j=1}^n {\alpha }_{j}E\left(I_{B_j}\mid A\right)\right)$

$=\sum _{j=1}^n {\alpha }_j\frac{E\left(I_B{I}_A\right)}{P\left(A\right)}$

$=\frac{1}{P\left(A\right)}E\left(\sum _{j=1}^n {\alpha }_j{I}_{B_j}{I}_A\right)$

$E(X\mid A)=\frac{E\left(X{I}_A\right)}{P_A}$

Then, the claim holds for all simple random variables. Now take any non-negative random variable$X$this carriable $X$can represent as the limit of non-negative, simple random

variables $X_n$

$\underset{n}{lim} X_n=Xa.s$

Now

$E(X\mid A)=E\left(\underset{n}{lim} X_n\mid A\right)$

$=\underset{n}{lim} \frac{E\left(X_n{I}_A\right)}{P\left(A\right)}$

$=\frac{E\left(\underset{n}{lim} X_n{I}_A\right)}{P\left(A\right)}$

$=\frac{E\left(X{I}_A\right)}{P\left(A\right)}$

Final Answer

We have to, take any random variable $X$and write it as$X=X^{+}-X^{-}$where $X^{*}=\max (X,0)$and $X^{-}=max(-X,0)$these random variables $X^{+}$and $X^{-}$therefore it is true that:

$E(X\mid A)=E\left(X^{+}-X^{-}\mid A\right)$

$=E\left(X^{+}\mid A\right)-E\left(X^{-1}\mid A\right)$

$=\frac{E\left(X^{+}{I}_A\right)}{P\left(A\right)}-\frac{E\left(X^{-}{I}_A\right)}{P\left(A\right)}$

$=\frac{E\left(\left(X^{+}-X^{-}\right)I_A\right)}{P\left(A\right)}$

$=\frac{E\left(\left(X\right)I_A\right)}{P\left(A\right)}$

Hence, the claim for all random variables is proved.