Covariance and Expectation of Discrete Random Variables

Question

I have to get back to basics here and need some help on checking some solutions. Given 2 discrete random variables $X$ and $Y$ and $\mathbb E (Y|X) = \mathbb E (Y) = \mu$ and $\mathbb E (X) = \lambda$.

(I) The covariance of $X$ and $Y$, denoted as $\mathbb C (X,Y)$ is

$$ \begin{align} \mathbb C (X,Y) &= \mathbb E[(X-\mu)(Y-\lambda)]\\ &=\mathbb E(XY) - \mu\lambda \end{align} $$

(II) Consider that both $X$ and $Y$ are independent. To find $\mathbb E (X|Y)$, $\mathbb E(Y|X)$ and $\mathbb C (X,Y)$, I have $$ \begin{align} \mathbb E (X|Y) = \mu\\ \mathbb E (Y|X) = \lambda\\ \mathbb C (X,Y) = \mathbb V(X) + \mathbb V(Y) \end{align} $$ Where $\mathbb V (A)$ is variance of $A$. What do you think?

Answer 1

By writing $E(Y\mid X) = E(Y)$ you essentially assume that the variables are "mean-independent" already. This in all cases implies that their covariance is zero (but not necessarily the reverse). By using the Law of Total Expectation $$E(XY) =E_X\left(E_{Y|X}(XY\mid X)\right) = E_X\left(XE_{Y|X}(Y\mid X)\right) = E_X\left(XE(Y)\right) = E(X)E(Y)$$

so zero covariance.

Answer 2

correct except the last one. $C(X,Y)= 0$. Note that this holds by independence, which implies $E(XY) = E(X)E(Y)$