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?)

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    $\begingroup$ Please pause to consider simple cases. For instance, suppose your population $\Omega$ contains a subset $E$ with a probability $p$ strictly between $0$ and $1.$ Define $Y$ to be the indicator of $E$ and $X=-Y.$ Thus, $F_X$ and $F_Y$ are piecewise constant; $F_X(x)$ jumps by $p$ at $-1$ and $1-p$ at $0,$ whereas $F_Y$ jumps by $1-p$ at $0$ and by $p$ at $1.$ As you may compute, $x(i)y(i)=0$ everywhere but the integrals of $x$ and $y$ are both nonzero. $\endgroup$ – whuber Sep 4 '18 at 13:53
  • $\begingroup$ @whuber But X is not independent of Y. Their correlation is -1. $\endgroup$ – luchonacho Sep 4 '18 at 13:59
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    $\begingroup$ Yes--but it is easy to construct independent variables with these two CDFs, whence the problem remains: what you are attempting to show obviously is not true. $\endgroup$ – whuber Sep 4 '18 at 14:14
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    $\begingroup$ You haven't a proof at all: you are invoking relationships among integrals that are not generally true. You will find that out by attempting to justify the manipulations you make: can you quote a property or theorem to support each one? $\endgroup$ – whuber Sep 4 '18 at 14:17
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    $\begingroup$ No, you are not: somehow you mysteriously turn a single integral into a double integral. $\endgroup$ – whuber Sep 4 '18 at 14:57

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,

$$ \int_0^1 x(i) y(i) \ \text{d}i = E(X) \ E(Y)$$

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    $\begingroup$ Please study my comment to your question. Your mathematical notation here, which has two "$\mathrm{d}i$" in the integral, makes no sense and may be leading you astray. $\endgroup$ – whuber Sep 4 '18 at 13:53

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