3
$\begingroup$

Let $\mu$ be a probability measure on $\mathbb R$ with Lebesgue density, i.e. $\mu(dx)=\mu(x)dx$. We aime to find increasing and decreasing functions $\phi_{+}: \mathbb R_+\to \mathbb R_{+}$ and $\phi_{-}: \mathbb R_+\to \mathbb R_{-}$ s.t. $\phi_{\pm}(0)=0$ and

$$\phi_{\pm}'(x)~~=~~\frac{1~-~F_{\mu}\big(\phi_+(x)\big)~+~F_{\mu}\big(\phi_-(x)\big)}{2\phi_{\pm}(x)\mu\big(\phi_{\pm}(x)\big)} \mbox{ for all } x>0,$$

where $F_{\mu}$ denotes the cumulative distribution function of $\mu$. My question is how to specify $(\phi_+,\phi_-)$ in terms of $\mu$? If there is some numerical shcema, I'm equally glad to know about it.

I find a solution if $\mu$ is symmetric, i.e. $\mu(x)=\mu(-x)$ for all $x\in\mathbb R$. Then it is easy to guess that $\phi_{\pm}=\pm\phi$ with

$$\phi^{-1}(x)~~=~~\int_0^x\frac{y\mu(y)}{1~-~F_{\mu}(y)}dy \mbox{ for all } x\in\mathbb R_+.$$

If someone knows how to treat the general case, please let me know. Thanks a lot!

$\endgroup$
  • 1
    $\begingroup$ The unknown functions in your ODE appear to be $\phi_{\pm}$. So I'm a bit confused by your question: how to specify the density $\mu$. In fact, it appears that $\mu$ is known or given or already specified. $\endgroup$ – Nawaf Bou-Rabee Oct 9 '16 at 13:12
  • $\begingroup$ I have reformulated my question. Here we assume that $\mu$ is regular enough. $\endgroup$ – CodeGolf Oct 9 '16 at 15:47
  • $\begingroup$ @NawafBou-Rabee Yes. Do you have some idea? $\endgroup$ – CodeGolf Oct 9 '16 at 15:49
  • $\begingroup$ Is the support of $\mu$ over all reals? Please specify all of your assumptions on $\mu$. Once we establish existence/uniqueness of a solution then we can approximate these ODEs. But I don't see that the right hand side is regular enough without more assumptions on $\mu$. $\endgroup$ – Nawaf Bou-Rabee Oct 9 '16 at 16:13
  • $\begingroup$ Generally, $\mu$ is supported on the whole real line. And in order to solve the system, the density $x\mapsto \mu(x)$ is assumed to be smooth. $\endgroup$ – CodeGolf Oct 9 '16 at 18:48
4
$\begingroup$

For simplicity, let's assume initially that the support of $\mu$ is the whole real line, so that $F$ is a homeomorphism $ \mathbb{R} \to (0,1) $. Let's denote $b:=F(0)=\mu(-\infty,0]=1-\mu[0,+\infty)\in(0,1)$.

The relevant function is the strictly convex function $\Phi :(0,1)\to\mathbb{R}$ defined as $$\Phi(t):=\int_{b}^t F^{-1}(s)\, ds $$ which is the Legendre transformation of $\int_0^x F(t)dt$. It has minimum value at $t=b$, and $\Phi(t)\to+\infty$, both for $t\to0$ and $t\to1$ (indeed $\Phi(0)=\int_{b}^0 F^{-1}(s)\, ds=\int_ {-\infty}^0 F(t) dt$: writing them in terms of the density function $\mu(s)$ as double integrals, and using Tonelli's theorem, one finds that both integrals diverge).

Let's denote $\Psi_{+}:=(\Phi_{|[b,1)})^{-1}: \mathbb{R}_+ \to[b,1)$ and $\Psi_-:=(\Phi_{|(0,b]})^{-1}:\mathbb{R}_+\to(0,b]$ . Note that $\Psi(t):=1-\Psi_+(t)+\Psi_-(t)=\big|\{\Phi\ge t\}\big|$ is the inverse function of the monotone decreasing rearrangement of $\Psi$.

Consider the solution $u:[0,+\infty)\to \mathbb{R}$ of the first order autonomous ODE

$$u'(t) = \frac{1}{2} \Psi (u ),\quad t\ge0$$ with $u(0)=0$, given explicitly by $$u^{-1}(x)=2\int_0^x\frac{dy}{ \Psi (y) }.$$

This integral is finite for all $x\ge0$ and diverges for $x\to+\infty$, so $u$ is a homeo $\mathbb{R}_+\to\mathbb{R}_+$ .

Then it is easy to check that $\phi_{\pm}:=F^{-1}\circ\Psi_\pm\circ u $ are defined $\mathbb{R}_+$, diverges to $\pm\infty$ for $x\to+\infty$, and solve the initial problem written as $$ 2\phi_+ F'(\phi_+)\phi_+'=2\phi_- F'(\phi_-)\phi_-'=1-F(\phi_+)+F(\phi_-)$$ that we can write equivalently, because $\phi_\pm=F^{-1}\circ F\circ \phi_\pm=\Phi'\circ F\circ \phi_\pm$, in the form $$ 2(\Phi\circ F\circ \phi_+)'=2(\Phi\circ F\circ \phi_-)'=1-\Psi_+\circ\Phi\circ F\circ \phi_+ +\Psi_-\circ\Phi\circ F\circ \phi_- .$$

Conversely, if $\phi_\pm$ solve the initial system, the latter equation shows that $\Phi\circ F\circ \phi_+=\Phi\circ F\circ \phi_-$ because they coincide at $0$ and have the same derivative, and in fact solve the above equation for $u$.

$$*$$

Rmk. The assumption on the support of $\mu$ is not crucial; if it is not the whole real line, $F$ is constant on some open set, and $F^{-1}$ has jumps and it is only defined as a left inverse of $F$; the integral for $\Phi$ is however well defined.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Why is $\mu(-\infty,0] = \mu[0,+\infty)$? $\endgroup$ – Nawaf Bou-Rabee Oct 9 '16 at 20:31
  • $\begingroup$ sorry, it was a typo, thank you, fixed. $\endgroup$ – Pietro Majer Oct 9 '16 at 20:34
  • $\begingroup$ That's an interesting transformation. $\endgroup$ – Nawaf Bou-Rabee Oct 9 '16 at 20:46
  • $\begingroup$ @ Pietro Majer Very nice solution. Thanks a lot! $\endgroup$ – CodeGolf Oct 10 '16 at 16:08
  • $\begingroup$ @ Pietro Majer Could you please specify a bit more about the definition of $\Psi$, i.e. ``$\Psi(t):=1-\Psi_+(t)+\Psi_-(t)=|\{\Psi\ge t\}|$ is the inverse function of the monotone decreasing rearrangement of $\Psi$''. Thanks a lot! $\endgroup$ – CodeGolf Oct 10 '16 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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