Question

Prove that$E\left[g\right(X)Y\mid X]=g\left(X\right)E[Y\mid X]$.

Short answer

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

Step-by-step solution

Given information

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

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

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

But, we have that

$f_{(X,Y)\mid X=x}(x,y)dxdy=f_{Y\mid X}(y\mid x)dy$

So the integral becomes

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

Which is equal to

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

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

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

Final answer

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