Expected value of a marginal distribution when the joint distribution is given I am asked to find the expected value of a vector of two random variables when the joint density is given. Is the recipe for solving this problem:


*

*Find the marginal distributions

*Find the expected values of the marginal distributions


which will involve a relatively long integration process, since I have to find the two marginals and then the two expected values?
Or is there some shortcut?
 A: As rightly pointed out by Dilip Sarwate, computing the expectation of one component as a two dimensional integral requires integrating out the other element of the vector:
$$\mathbb{E}[X] =   \iint x\,f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx,$$
(which is a special case of the so-called law of the uncounscious statistician). The only simplifications I can think of is

*

*when finding the conditional expectation of one component given the other is easier:
$$\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X|Y]] =  \int x\,f_{X|Y}(x|y)\,\mathrm dx\,f_Y(y)\,\mathrm dy,$$
by the double projection theorem, in the sense that this could require computing a single marginal instead of two;

*when computing the marginal cdf is easier, since
$$\mathbb{E}[X] = \int_{-\infty}^0 F_X(x)\,\mathrm dx-\int_{+\infty}^0 (1-F_X(x))\,\mathrm dx$$
by an integration by parts.

A: Using Monte Carlo estimate, the expected value for the marginal distribution will the $\frac{\displaystyle\sum_{i=1}^Nx_i}{N}$ where $x_i$ are x's samples from the (x,y) samples from joint distribution.
By Monte Carlo integration, $\int\int\phi(x,y)f(x,y)dxdy$ can be estimated by $\frac{1}{N}\displaystyle\sum_{i=1}^N\phi(x_i,y_i)$. Here, we have $\phi(x,y)= x$. 
