Expected value of $g(x, y)$ without having their joint distribution For two random variables: $X$, $Y$. If their marginal distributions are given $f_X(x)$, $f_Y(y)$ and $g(x, y)$ is some function of $X$ and $Y$. Can I get the expected value of $g(x, y)$ if I know their covariance matrix but I do not have their joint distribution $f_{X,Y}(x, y)$
 A: Here's an explicit counterexample:$\DeclareMathOperator{\E}{\mathbb E}$
Consider the function $g(x, y) = x^2 y^2$ and the marginal distributions $X \sim \mathcal N(0, 1)$, $Y \sim \mathcal N(0, 1)$, and no covariance: $\E[X Y] = 0$.
If $X$ and $Y$ are independent, so that $\begin{bmatrix}X \\ Y\end{bmatrix} \sim \mathcal N\left( \begin{bmatrix}0 \\ 0\end{bmatrix}, \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \right)$,
then
$$
\E\left[g(X, Y)\right]
= \E\left[X^2 Y^2\right]
= \E\left[X^2\right] \E\left[Y^2\right]
= 1
.$$
But consider instead the case $Y = Z X$, where $Z \sim \operatorname{Uniform}(\{-1, 1\})$ is independent.
We can check that the marginal distribution of $Y$ is still correct:
\begin{align}
     \Pr(Y \le y)
  &= \Pr(Z X \le y)
\\&= \Pr(Z = 1) \Pr(X \le y \mid Z = 1) + \Pr(Z = -1) \Pr(X \ge -y \mid Z = -1)
\\&= \frac12 \Pr(X \le y \mid Z = 1) + \frac12 \Pr(X \ge -y \mid Z = -1)
\\&= \frac12 \Pr(X \le y) + \frac12 \Pr(X \ge -y)
\\&= \Pr(X \le y)
\end{align}
because $Z$ is independent of $X$ and the distribution of $X$ is symmetric about 0.
We also have that $X$ and $Y$ are uncorrelated:
$$
\E[X Y] - \E[X] \E[Y]
= \E[X (Z X)] - 0
= \E[X^2] \E[Z]
= 1 \cdot 0
= 0
.$$
But the expectation of $g$ is different than the other case:
$$
\E\left[g(X, Y)\right]
= \E\left[X^2 Y^2\right]
= \E\left[X^2 (Z X)^2\right]
= \E\left[X^4\right] \E\left[Z^2\right]
= 3
.$$

Here's the one case where you can compute it:
suppose that $g$ is of the form
$$g(x, y) = g_x(x) + g_y(y) + \alpha x y + \beta.$$
Then
$$
\E[g(X, Y)]
= \E[g_x(X)] + \E[g_y(Y)] + \alpha \E[X Y] + \beta
,$$
so you can compute the expectations of $g_x$ and $g_y$ using the marginal distribution, and you know $\E[X Y] = \rho + \E[X] \E[Y]$.
A: There may be special cases, but in general, If you could get any $E[g(x,y)]$, then you could get any joint moment. And, that should reveal the joint distribution, and I don't think you can do this using only covariance matrix.
Thinking in 1D, if you have any moment, you can write the moment generating function, which lead to characteristic function, which also lead to PDF. In 2D, if you have any joint moment, that should in some way similarly lead to joint PDF.
