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Q.26

Expert-verifiedFound in: Page 361

Book edition
9th

Author(s)
Sheldon M. Ross

Pages
432 pages

ISBN
9780321794772

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

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

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

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

$E\left(g\right(X)Y\mid X=x)={\iint}_{{\mathrm{\mathbb{R}}}^{2}}\u200ag\left(x\right)y{f}_{(X,Y)\mid X=x}(x,y)dxdy$

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

${\int}_{\mathrm{\mathbb{R}}}\u200ag\left(x\right)y{f}_{Y\mid X}(y\mid x)dy$

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

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

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

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