Pushforward-change-of-variable with quantile function

Question

I’ve been dealing with an issue about change-of-variable formula.

Let $\mu$ be a probability measure on $\mathbf R_+$. Let $F(x) = μ([0,p])$ and $Q$ its quantile function, ie $Q(p) = \inf \{q \in \mathbf R_+ : F(q) \ge p\}$. As $Q$ is increasing, it is Borel.

Thus, let $\ell$ be Lebesgue measure on $\mathbf R_+$ and $Q_*(\ell)$ its pushforward by $Q$, ie $Q_*(\ell)(A) = \ell(Q^{-1}(A))$. It is well-known (Galois formula) that for all $x \in \mathbf R_+$ and $p \in [0,1]$, $Q(p) \le x \iff p \le F(x)$. Thus, if $0 \le a \le b$, then $$Q_*(\ell)((a,b]) = \ell(Q^{-1}( (a,b])) = \ell((F(a), F(b)]) = F(a) - F(a) = \mu((a,b])$$

The familiy of intervals of form $(a,b]$ is a $\pi$-system on the set $\mathbf R_+$ which generates the Borel $\sigma$-algebra $\mathscr B(\mathbf R_+)$. It follows that $Q_*(\ell)$ and $\mu$ are the same measure.

One can define the Lorenz function : $λ(p) = \int_0^p Q(t) \ell(\mathrm d t)$. Now, I apply the change-of-variable formula, and I get : $$ \lambda(p) = \int_{Q(0)}^{Q(p)} u~ Q_*(\ell)(\mathrm d u) = \int_0^{Q(p)}u \mu(\mathrm d u) $$

But this equality is false if (and only if ?) $\mu$ has atoms. The simplest possible example is $\mu = \delta_1$, the Dirac mass in $1$ : $Q$ is a function that maps $0$ to $0$ and any value $p \in (0,1]$ to $1$, so $\lambda$ is the identity function on $[0,1]$. But $\int_0^{Q(p)}u \mu(\mathrm d u)$ is the same function than $Q$.

Then, two questions :

1. What is wrong in the previous calculus?
2. Is the equality $\lambda(p) = \int_0^{Q(p)} u \mu(\mathrm d u)$ true if one assumes $μ$ is diffuse (ie has no atom)?

Thanks in advance for your help!

Answer 1

Two years after, and as I read someone making the same mistake than me, I make a detailed answer.

The mistake lies in the way to apply the pushforward change of variable formula. Indeed, a general version that can be found in handbooks (or Wikipedia) is:

Let two measurable spaces $(E, \Sigma_1), (F, \Sigma_2)$, a measure $\nu$ on $(E, \Sigma_1)$ and two measurable mappings $f:E \longrightarrow F$ and $g: F \longrightarrow \mathbf R$.

The function $g \circ f$ is integrable with respect to the measure $\nu$ if, and only if the function $g$ is integrable with respect to the measure $f_*\mu$. In that case,

$$\int_{X_1} g \circ f~\mathrm{d}\nu = \int_{X_2} g~\mathrm{d}(f_*\nu).$$

Here, the goal is to apply the theorema with $\nu = \ell$ the Lebesgue measure, $f=Q$ and $g$ the identity function. Then as explained in the question, $Q_*\ell = \mu$.

But what do we chose for $X_1$ and $X_2$? My mistake lies in the fact that I took $X_1 = [0,p]$ and $X_2 = [0, Q(p)]$. But then, $f$ is not the function $Q$ anymore, but the restriction of $Q$ to the domain $[0,p]$ and the codomain $[0,Q(p)]$. This innocent mistake may make the equality $Q_*\ell = \mu$ fail!

Actually, the correct way to perform the change-of-variable formula on a subset is to add an indicator function in the integrand. I.e., we set:

• $X_1 = [0,1]$;
• $X_2 = \mathbf R_+$;
• $\nu = \ell$, the Lebesgue measure on $[0,1]$;
• $f = Q: [0,1] \longrightarrow \mathbf R_+$;
• $\begin{array}{rcrcl}g&:& \mathbf R_+ & \longrightarrow & \mathbf R_+\\ && x & \longmapsto & x \mathbf 1_{[0, Q(p)]}(x).\end{array}$

Then, we can apply the pushforward formula: $$ \int_{[0,1]} g(Q(u))~\mathrm d\ell(u) = \int_{\mathbf R_+} g(x)~\mathrm d(Q_*\ell)(x). $$

The right-hand integral makes no difficulty. Using the fact that $Q_*\ell = \mu$, it can directly be rewritten: $$\int_{\mathbf R_+} x \mathbf 1_{[0,Q(p)]}(x)~\mathrm d\mu(x) = \int_0^{Q(p)} x~\mathrm d \mu(x).$$

But on the right-hand side, we now have to deal with the $\mathbf 1_{[0,Q(p)]}(Q(u))$, which is clearly not a mere $\mathbf 1_{[0,p]}(u)$. Actually, the Galois inequalities implies that $Q(u) \le Q(p) \iff x \le F(Q(u))$. Thus, the right-hand integral writes as: $$\int_{\mathbf [0,1]} Q(u) \mathbf 1_{[0,Q(p)]}(Q(u))~\mathrm d\ell(u) = \int_{\mathbf [0,1]} Q(u) \mathbf 1_{[0,F(Q(p))]}(u)~\mathrm d\ell(u) = \int_0^{F(Q(p))} Q(u)~\mathrm d\ell(u).$$

So, the correct result is: $$\int_0^{F(Q(p))} Q(u)~\mathrm d\ell(u) = \int_0^p x~\mathrm d\mu(x). $$

Note that $F(Q(p)) = p$ as soon as $\mu(\{Q(p)\}) = 0$. Furthermore, $Q(p) > 0$ iff $p > \mu(\{0\})$. Thus:

• Either $\mu$ has no atoms except 0, and the equality $\lambda(p) = \int_0^{Q(p)} x~\mathrm d\ell(x)$ stands for every $p$.
• Either $\mu$ has atoms in values other than 0, and the equality fails for some values. To be more precise, it fails for every values of $p$ such that $F(Q(p)) \neq p$ and $Q(p) \neq 0$. Graphically, this corresponds to the cases where $F$ “jumps” over the value $p$.

---

The lesson of this is: do not apply the pushforward change-of-variable formula with a restricted function unless you’re absolutely sure on what is happening. In case of doubt, prefer adding an indicator function into your integrand.