Question

Show that if $X$and $Y$are independent, then

$E[X\mid Y=y]=E\left[X\right]\text{for all}y$

(a) in the discrete case;

(b) in the continuous case.

Short answer

The calculation for both cases is similar. Just use the fact that $X$ and $Y$ are independent.

Step-by-step solution

Given information(part a)

Given in the question that,$E[X\mid Y=y]=E\left[X\right]\text{for all}y$

Explanation (part a)

Let's begin from the left side. We have that

$E(X\mid Y=y)=\sum _x xP(X=x\mid Y=y)$

$=\sum _x xP(X=x)$

$=E\left(X\right)$

where the second equality is the consequence of the fact that $X$ and $Y$ are independent.

Final answer(part a)

We proved that $X$and $Y$are independent

Given information(part b)

Given in the question that,$E[X\mid Y=y]=E\left[X\right]\text{for all}y$

Explanation(part b)

Let's again start from the left side. For $y\in supp{f}_Y$, we have that

$E(X\mid Y=y)={\int }_{\mathrm{ℝ}} x{f}_{X\mid Y}(x\mid y)dx$

Because of the independence, we have that

$f_{X\mid Y}(x\mid y)=\frac{f(x,y)}{f_Y\left(y\right)}=\frac{f_X\left(x\right)f_Y\left(y\right)}{f_Y\left(y\right)}=f_X\left(x\right)$

so we have that the integral is equal to

${\int }_{\mathrm{ℝ}} x{f}_{X\mid Y}(x\mid y)dx={\int }_{\mathrm{ℝ}} x{f}_X\left(x\right)dx=E\left(X\right)$

so we have demonstrated the asserted in the two cases.

Final answer(part b)

We proved that $X$ and $Y$ are independent