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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

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:

$E(X鈭�A)=E_a\left(I_a\right)$

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

$=P_A\left(B\right)$

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

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

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

$X=\underset{j=1}{\overset{n}{鈭�}}{伪}_j{I}_n$

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

$E(X鈭�A)=E\left(\underset{j=1}{\overset{n}{鈭�}}鈥�{伪}_j{I}_{B_j}鈭�A\right)$

$=\underset{j=1}{\overset{n}{鈭�}}鈥�{伪}_j\frac{E\left(I_B{I}_A\right)}{P\left(A\right)}$

$=\frac{1}{P\left(A\right)}E\left(\underset{j=1}{\overset{n}{鈭�}}鈥�{伪}_j{I}_{B_j}{I}_A\right)$

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 $X_n$

$\underset{n}{lim}鈥�X_n=Xa.s$

Now

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

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

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

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

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

$E(X鈭�A)=E\left(X^{+}鈭�X^{鈭�}鈭�A\right)$

$=E\left(X^{+}鈭�A\right)鈭�E\left(X^{鈭�1}鈭�A\right)$

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

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

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

Hence, the claim for all random variables is proved.