# 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?)

• 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. – whuber Sep 4 '18 at 13:53
• @whuber But X is not independent of Y. Their correlation is -1. – luchonacho Sep 4 '18 at 13:59
• 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. – whuber Sep 4 '18 at 14:14
• 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? – whuber Sep 4 '18 at 14:17
• No, you are not: somehow you mysteriously turn a single integral into a double integral. – 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,

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

• 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. – whuber Sep 4 '18 at 13:53