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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!

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  • $\begingroup$ Okay, I eventually solved it: the integration bounds in the right-hand integral are not Q(0), Q(p) if Q is not strictly increasing, ie μ is not diffuse. $\endgroup$ – Valentin Melot Aug 22 '18 at 21:10

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