If $X$ has mean $\mu_x$, variance $\sigma_x^2$, and $Y$ has mean $\mu_y$, variance $\sigma_y^2$, what are the mean and cov of $(X, Y)$?

Question

If r.v. $X$ has mean $\mu_x$, variance $\sigma_x^2$, and r.v $Y$ has mean $\mu_y$, variance $\sigma_y^2$. I'm interested in the expectation and covariance of the random vector $\begin{bmatrix} X \\ Y \end{bmatrix}$. Without any additional information about the joint distribution, is it always true that

$$\mathbb{E}\begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix}$$ $$Cov\begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \sigma_x^2 & \sigma_{xy} \\ \sigma_{xy}& \sigma_y^2 \end{bmatrix}$$

where $\sigma_{xy}$ is the covariance $Cov(X, Y)$?

Answer 1

Yes, the formula is true by the definitions of expectation of random vectors and covariance matrices.

But when you calculate $E(XY)$, which is needed when you calculate the covariance $\textrm{Cov}(X,Y)$, you need information for the joint distribution.

Answer 2

Yes, the expectation is calculated elementwise, so the expected value of the random vector $(X, Y)^T$ it is simply column vector of $\mu_x$ and $\mu_y$, i.e., $\mu = (\mu_x, \mu_y)^T$. For the covariance, denote $Z = (X, Y)^T$, thus by definition $$ Cov(Z) = E[(Z-\mu)(Z-\mu)^T] $$ write $(Z-\mu)(Z-\mu)^T$ explicitly $$ (Z-\mu)(Z-\mu)^T = \begin{pmatrix} (X-\mu_x)(Y-\mu_y) \quad (X-\mu_x)(Y-\mu_y)\\ (Y-\mu_y)(X-\mu_x) \quad (Y-\mu_y)(Y-\mu_y) \end{pmatrix}, $$ and then apply expectation elementwise, so you got for the entries $\Sigma_{12}$ , and $\Sigma_{21}$ the definition of covariance between $X$ and $Y$, i.e., $E[(X-\mu_x)(Y-\mu_y)] = Cov(X,Y) = \sigma_{xy}$.