In the following problem, $X$ and $Y$ are continuous random variables with joint density
$$f(x,y)=\frac{3}{4}(2-x-y)\cdot\mathsf 1_{(0,2)^3}(x,y,x+y)$$
An insurance policy pays total $X+Y$. Find expected amount paid.
Ok, so there's the problem. The solution goes on to explain that you can answer this simply by finding the marginal distribution of either $X$ or $Y$ and then taking the expectation of either one of those. The answer is $1/2$. The solution itself was easy enough to execute, but my main question is: Why is this ok to do? That is, under what circumstance can I assume that, in seeking to find the expected value of a joint distribution, can I do a marginal and take the expected value from there?
My first instinct was to do LOTUS with double integral, but this ended up being wrong and taking forever with very tedious calculus...
While we can compute the marginal densities and sum $\mathbb E[X]$ and $\mathbb E[Y]$, it is simpler in this case to compute the expectation directly. Let $g:\mathbb R^2\to\mathbb R$ be defined by $g(x,y) = x+y$, then as $g$ is a (Lebesgue) measurable function, by the law of the unconscious statistician we have \begin{align} \mathbb E[X+Y] &= \int_{(0,2)^3} g(x,y)f_{X,Y}(x,y)\ \mathsf d(x\times y)\\ &=\int_0^2\int_0^{2-x} (x+y)\frac34(2-x-y)\ \mathsf dy\ \mathsf dx\\ &= 1. \end{align}
Computing the marginal densities, we have for $0<x<2$: $$ f_X(x) = \int_0^{2-x}\frac34(2-x-y)\ \mathsf dy = \frac{3}{8} (x-2)^2 $$ and hence $$ \mathbb E[X] = \int_0^2\frac{3}{8}x (x-2)^2\ \mathsf dx = \frac12. $$ The same computation with $X$ and $Y$ transposed yields $$ f_Y(y) = \frac{3}{8} (y-2)^2\cdot\mathsf 1_{(0,2)}(y), $$ and hence $\mathbb E[Y] = \frac12$, so that indeed $$ \mathbb E[X+Y] = \mathbb E[X] + \mathbb E[Y] = \frac12+\frac12 = 1. $$