Let $X$ be a standard normal random variable and $Y = X^2$.

Question

Let $X$ be a standard normal random variable and $Y = X^2$.

Prove that $E[Y|X] = k$, where $k$ is a constant, implies $Cov(X,Y) = 0$.

Thanks

Answer 1

$$Cov (X,Y)= E(XY)-E(X)(Y)$$ Since X is standard normal then $X^2$ follows chi-square with 1 degrees of freedom.This tells you that E(Y)=1 Using the law of total expectation i.e $$E[E[Y|X=x]]=E[Y]$$ Since $$XY=X^3$$ and since X is a symmetric random variable the odd expectations are zero.

Answer 2

In this answer, we will show that, whenever we have random variables $X,Y$ s.t. $\mathbb{E}[Y|X]=k$, then their covariance will be zero.

For the answer, I will use the properties of conditional expectation on a random variable. You can view these properties in page 4 in this link. Using properties in the link I mentioned, we deduce

1. $Cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$.
2. $\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]=k$. (properties $(i)$ and $(v)$)
3. $\mathbb{E}[XY]=\mathbb{E}[\mathbb{E}[XY|X]]=\mathbb{E}[X\mathbb{E}[Y|X]]=\mathbb{E}[kX]=k\mathbb{E}[X]$. (properties $(v)$ and $(vi)$)

Using 1, 2 and 3 we conclude the result.