Expected value of joint quantile functions Consider a population of individuals (not a sample). We are interested in two variables, $X$ and $Y$, which are independent. $X$ distributed with pdf $f(x)$ and CDF $F(x)$, and $Y$ distributed with pdf $g(y)$ and CDF $G(y)$. 
Consider the quantile funtions of these two distributions. Importantly, the support of these two functions is the same, because the variables belong to the same population. Thus, we have:
$$ x(i) = Q_x(i) =  F^{-1}(i) \qquad y(i) = Q_y(i) = G^{-1}(i) \tag{1}\label{1}
$$
(You can think of this set up in the alternative way. Assign a continuous index $i \in [0,1]$ to every person in the (infinite) population such that their values of $X$ and $Y$ correspond to the quantile functions above.)
Now, I have the following term:
$$ \int_0^1 x(i)  y(i) \; \text{d}i \tag{2}\label{2}
$$
Since the integral of the quantile function over the whole support is equal to the mean (see this answer), I get the idea that somehow the above term is equivalent to
$$ \int_0^1 x(i) \ \text{d}i \int_0^1 y(i) \ \text{d}i  \tag{3}\label{3}
$$
which is equivalent to the multiplication of the means. 
In other words, I am looking for the equivalent of $$E(XY)=E(X)E(Y) \tag{4}\label{4}$$ using quantile functions.
Below is how far I've got. Let's start from the definition of expected value for independent random variables:
$$ \int_X \int_Y xy \; f(x) g(y) \; \text{d}x \; \text{d} y = E(X) \; E(Y)  \tag{5}\label{5}
$$ 
Now, implicit differentiation means that:
$$ \text{d}i = \frac{\partial F(x)}{\partial x} \text{d}x = f(x) \text{d}x\qquad \text{d}i = \frac{\partial G(y)}{\partial y} \text{d}y = g(y) \text{d}y  \tag{6}\label{6}
$$
Replacing these above we get to:
$$ \int_0^1 \int_0^1 x(i) y(i) \; \text{d}i \; \text{d} i = E(X) \; E(Y)  \tag{7}\label{7}
$$ 
Here I'm stuck. Any ideas how to proceed (if it is actually possible?)
 A: Quite noob indeed. Notice that:
$$ \int_0^1 \int_0^1 x(i) y(i) \ \text{d}i \ \text{d} i = E(X) \ E(Y)$$ 
is also
$$ \int_0^1 \left( \int_0^1 x(i) y(i) \ \text{d}i \right) \text{d} i = E(X) \ E(Y)$$ 
But the term inside the parenthesis, whatever it is, does not depend on $i$. Denote it as $C$, then:
$$ \int_0^1 C \ \text{d} i = E(X) \ E(Y)$$ 
Solving the integral, we get:
$$ \int_0^1 C \ \text{d} i = C i \Big|^1_0 = C = E(X) \ E(Y)$$ 
This is,
$$ \int_0^1 x(i) y(i) \ \text{d}i = E(X) \ E(Y)$$ 
A: If the integrand $f(x,y)$ can be partitioned into a product of functions of the individual differentials $$f(x,y) = g(x) h(y)$$  then we can write a double integral as a product of single integrals.
$$ \int_{a}^{b}\int_{c}^{d} g(x) h(y) dx dy = \left(\int_{c}^{d} g(x) dx\right)\left(\int_{a}^{b}  h(y) dy\right)$$

Now, I have the following term:
$$ \int_0^1 x(i)  y(i) \; \text{d}i 
$$

This makes little sense, but possibly you meant to write
$$ \int_0^1 \int_0^1 x(p_1)  y(p_2) \; \text{d}p_1\text{d}p_2
$$
