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Expert-verified Found in: Page 361 ### A First Course in Probability

Book edition 9th
Author(s) Sheldon M. Ross
Pages 432 pages
ISBN 9780321794772 # Prove that $E\left[g\left(X\right)Y\mid X\right]=g\left(X\right)E\left[Y\mid X\right]$.

We prove that, $E\left[g\left(X\right)Y\mid X\right]=g\left(X\right)E\left[Y\mid X\right]$

See the step by step solution

## Step 1: Given information

Given in the question that, we have to prove that $E\left[g\left(X\right)Y\mid X\right]=g\left(X\right)E\left[Y\mid X\right]$

## Step 2: Explanation

Because of the clarity, suppose that $X$ and $Y$ are continuous random variables with its density functions. Take any $x\in$ $supp{f}_{X}$. We have that

$E\left(g\left(X\right)Y\mid X=x\right)={\iint }_{{\mathrm{ℝ}}^{2}} g\left(x\right)y{f}_{\left(X,Y\right)\mid X=x}\left(x,y\right)dxdy$

But, we have that

${f}_{\left(X,Y\right)\mid X=x}\left(x,y\right)dxdy={f}_{Y\mid X}\left(y\mid x\right)dy$

So the integral becomes

${\int }_{\mathrm{ℝ}} g\left(x\right)y{f}_{Y\mid X}\left(y\mid x\right)dy$

Which is equal to

$g\left(X\right)E\left(Y\mid X=x\right)$

Since the relation hold for every $x\in suppf$, we end up with

$E\left(g\left(X\right)Y\mid X\right)=g\left(X\right)E\left(Y\mid X\right)$

We prove that, $E\left[g\left(X\right)Y\mid X\right]=g\left(X\right)E\left[Y\mid X\right]$ ### Want to see more solutions like these? 