Q.27 - A First Course in Probability

Question

Prove that if $E[Y\mid X=x]=E\left[Y\right]$ for all $x$, then $X$ and $Y$ are uncorrelated; give a counterexample to show that the converse is not true.

Hint: Prove and use the fact that $E\left[XY\right]=E\left[XE\right[Y\mid X\left]\right]$.

Step-by-Step Solution

Answer

We prove that,$E[Y\mid X=x]=E\left[Y\right]$ for  all $x$

Step 1: Given information

Given in the question that, we have to prove that $E[Y\mid X=x]=E\left[Y\right]$ for all $x$

Step 2: Explanation

Let's find the covariance between $X$ and $Y$. We have that

$\mathrm{Cov}(X,Y)=E\left(XY\right)-E\left(X\right)E\left(Y\right)$

Since we have that $E(Y\mid X=x)=E\left(Y\right)$for all $x$es, we have that

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

Which implies

Hence

$\mathrm{Cov}(X,Y)=0$

Step 3: Disapprove the converse

Now, let's disapprove the converse. Consider random variable $X~\mathcal{N}(0,1)$ and define $Y=X^2$. We have that

$\mathrm{Cov}(X,Y)=\mathrm{Cov}\left(X,X^2\right)=E\left(X^3\right)-E\left(X\right)E\left(X^2\right)$

Observe that $X^3$ is symmetric around zero, so $E\left(X^3\right)=0$. Finally, we see that $C o v(X, Y)=0$. But, we have that

$E(Y\mid X)=E\left(X^2\mid X\right)=X^2$

which is not obviously equal to constant $E\left(Y\right)=E\left(X^2\right)$.

Step 4: Final answer

We prove that, $E[Y\mid X=x]=E\left[Y\right]$ for all $x$

And from the above example we prove that $E\left(Y\right)=E\left(X^2\right)$