Intuition behind $E(XY) = E(X) E(Y) $ for independent random variables $X,Y$

Question

I have been wondering what's the intuition behind a well known result: $E(XY) = E(X) E(Y) $ for independent random variables $X,Y$

I found this post: here which kinda solves the problem.

But, the explanation given there seems to be not clear enough for me.

What I think: Without loss of generality, we know that besides independence we can assume that both random variables, $X$ and $Y$ are simple random variables, and so, it is possible to represent them as, i.e. taking X first:

$X = \sum^n_{i=1} a_i 1_{A_i}$, then compute the product $XY$ and take expectation.

But could somebody please explain the intuition behind it to me?

I really want to get the notion of how to understand the result of that post (which i believe is correct)

Thank you all guys.!

Answer 1

It is hard to give precise answer since you are asking for intuition.

Suppose for a certain number b you will compute bX. What’s the expected value of this computation? Well, if the realization of the variable X was done independently of the choice of the number b, then your computation will produce on average b.EX. Now make b random...

Answer 2

We know that

$E[XY]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xyf_{XY}(x,y)dxdy$

If the RVs are independent then

$E[XY]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xyf_{X}(x)f_{Y}(y)dxdy \\ \qquad \quad=\int_{-\infty}^{\infty}xf_{X}(x)dx\int_{-\infty}^{\infty}yf_{Y}(y)dx \\ \qquad \quad = E[X]E[Y]$

For the two jointly distributed Gaussian $X,Y$ RVs it means uncorrelatedness $(\rho_{xy}=0,\Sigma = \mathrm{diag}\{\sigma_{x}^2,\sigma_{y}^2\})$ $(f_{XY}(x,y)=f_{X}(x)f_{Y}(y))$ implying independence (Not true for other densities). But it tells us that one variable does not contain information about the other random variable.