14
$\begingroup$

I was wondering where there is a general formula to relate the expected value of a continuous random variable as a function of the quantiles of the same r.v. The expected value of r.v. $X$ is defined as:
$E(X) = \int x dF_X(x) $ and quantiles are defined as : $Q^p_X = \{x : F_X(x) = p \} =F_X^{-1}(p) $ for $p\in(0,1)$.

Is there for instance a function function $G$ such that: $E(X) = \int_{p\in(0,1)} G(Q^p_X) dp $

$\endgroup$

1 Answer 1

24
$\begingroup$

The inverse (right inverse in discrete case) of the cumulative distribution function $F(x)$ is called the quantile function, often denoted $Q(p)=F^{-1}(p)$. The expectation $\mu$ can be given in terms of the quantile function (when the expectation exists ...) as $$ \mu=\int_0^1 Q(p)\; dp $$ For the continuous case, this can be showed via a simple substitution in the integral: Write $$ \mu = \int x f(x) \; dx $$ and then $p=F(x)$ via implicit differentiation leads to $dp = f(x) \; dx$: $$ \mu = \int x \; dp = \int_0^1 Q(p) \; dp $$ We got $x=Q(p)$ from $p=F(x)$ by applying $Q$ on both sides.

For the general case, we can interpret $E(X) = \int x dF_X(x)$ as a Riemann–Stieltjes integral which, when the range of $X$ is the finite interval $[a,b]$, is defined as a limit of approximating sums of the form, for partitions $a=x_0<x_1<\dotsm<x_n=b$, $$\sum_{i=0}^{n-1} x_i^* \left[F(x_{i+1})-F(x_i)\right] $$ Passing to the quantile function $F^*$ (read the $^*$ as indicating a generalized inverse), and using $p_i=F(x_i)$, $p_i^*=F(x_i^*)$ this goes over to the approximating sum $$ \sum_{i=0}^{n-1} F^*(p_i^*) \left[ p_{i+1} - p_i \right] $$ which are approximating sums for $$ \int_0^1 F^*(p) \; dp $$ Then the infinite range case is treated the same way as with Riemann integrals.

$\endgroup$
1
  • 1
    $\begingroup$ Can you have a look at this question please? I think your insights might be helpful. $\endgroup$
    – luchonacho
    Commented Sep 4, 2018 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.