Expected value (Mean) of a joint distribution

Question

I saw in a textbook that if we have a joint distribution $f(X,Y)$ that is a Gaussian distribution, then we have the mode equal to the mean. The mode is just the values of $X$ and $Y$ such that $f(X,Y)$ is maximum. I can see how the mode is the same as the mean with a graph. However, I have a lot of trouble understanding how to formulate the mean mathematically in this case. How do I calculate the expected value using an integral? What is $E(X,Y)$? I have never seen it before.

I suppose in a discrete case, it would just be $p_1[X_1, Y_1]' + p_2[X_2,Y_2]'$ (assuming there are only two possible values for $X,Y$) but I don't know how to transform this into continuous case.

Answer 1

If $(X,Y)$ is a random vector, then the mean of the random vector is defined as $(E[X], E[Y])$, so you just need to take the mean of each component separately. If $(X,Y)$ has a joint density $f$, then for instance $$E[X] = \int_{-\infty}^\infty\int_{-\infty}^\infty x f(x,y) \, dx \, dy.$$ Note that in cases where $X$ and $Y$ are independent, the joint density decomposes as $f(x,y) = g(x) h(y)$ into the marginal densities, so the above integral would just be the usual formula $E[X] = \int_{-\infty}^\infty x g(x) \, dx$.