Variance and covariance problem

Question

I just got stuck with an exercise, see if you could give me a hand.

Let $X$ and $Y$ be two random variables with two-dimensional joint distribution. Knowing that $E(X)=E(Y), \ E(X^2) = E(Y^2)$ and $Cov(X,Y) = 2$, reason if it is true or false that $Var(X) < 2$.

I have tried to see if there is some way to bound the variance (knowing $Var(X)=Var(Y)$ because of equality of expectations) or if it can be seen through the covariance expression, but I am getting nowhere.

Thanks a lot.

Answer 1

Using the Cauchy–Schwarz inequality, $$ 2=|\operatorname{Cov}(X,Y)|\le \sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}=\operatorname{Var}(X). $$

Answer 2

I'm also getting that $\mathbb{Var}(X)\geq 2$.

First of all, $$\begin{eqnarray*}2&=&\mathbb{Cov}(X,Y) \\&=& \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y) \\&=& \mathbb{E}(XY)-\left(\mathbb{E}(X)\right)^2\end{eqnarray*}$$ This implies $$\mathbb{E}(XY)=\left(\mathbb{E}(X)\right)^2+2$$ Hence, $$\begin{eqnarray*}\mathbb{E}\left((X-Y)^2\right) &=&\mathbb{E}(X^2)+\mathbb{E}(Y^2)-2\mathbb{E}(XY) \\ &=& 2\left[\mathbb{E}(X^2)-\left(\mathbb{E}(X)\right)^2\right]-4 \\&=& 2\mathbb{Var}(X)-4\end{eqnarray*}$$ But $\mathbb{E}\left((X-Y)^2\right)\geq 0$ which means $\mathbb{Var}(X)\geq 2$.