time derivative of the median of a stochastic process

Question

Suppose you have a cumulative distribution that is changing with time, namely $ P_t(x) $. Assume $ P_t $ is monotone increasing and smooth enough so that we can define $ x_t(P) = P_t^{-1} $. We want to track percentiles over time, so that the rate of change of the median would be described by \begin{equation*} \frac{d}{dt} x_t(1/2). \end{equation*} We could define a kind of fixed point (``fixed percentile''?) of the time evolution in the following sense: $ P_\ast $ is a fixed percentile at time $ t $ if it satisfies \begin{equation*} \frac{d}{dt} x_t(P_\ast) = 0. \end{equation*}

My question is: Does this ``fixed percentile'' have an established or better name? Is it used and defined in the literature? Do we know any of its properties, such as under what conditions it will exist and/or be unique?

For a concrete example, consider the Pareto distribution with scale parameter $ \beta(t) $ and shape parameter $ \alpha(t) $ so that \begin{equation*} P_t(x) = 1 - \left( \frac{\beta}{x} \right)^\alpha, \end{equation*} with $ x \geq \beta $. Then \begin{equation*} x_t(P) = \frac{\beta}{(1-P)^{1/\alpha}} \end{equation*} and \begin{equation*} \frac{dx_t}{dt} = \frac{\beta' \alpha^2 + \beta \alpha' \ln (1-P)}{\alpha^2 (1-P)^{1/\alpha}} \end{equation*} Solving for a fixed percentile, we get \begin{equation*} P_\ast = 1 - \exp\left(\frac{-\beta' \alpha^2}{\beta \alpha'}\right). \end{equation*} To make it even more concrete, let $ \beta(t) = t $ and \begin{equation*} \alpha = \frac{-\ln (2)}{\ln (t) - 5} \end{equation*} then $ P_\ast = 1/2 $ is independent of time ($ 1 \leq t < e^5 $), and the distribution is moving to the right ($ \beta' > 0 $) while simultaneously spreading out ($ \alpha' < 0 $) so that the median remains constant. In this case the median is a repelling fixed point of the distribution, separating the rich getting richer from the poor getting poorer.

Answer 1

Here's another class of examples. Let's assume that $P_t(x)=\int_{-\infty}^xp(y,t)dy$, where $p(y,t)$ is a solution to the Fokker-Planck equation $$\partial_tp(y,t)=-\partial_y\big(f(y)p(y,t)\big)+\frac12\partial_y^2\big(g^2(y)p(y,t)\big).$$ By your notation, we have $$P_{\ast}=\int_{-\infty}^{x_t}p(y,t)dy.$$ Hence, $$\frac{dP_{\ast}}{dt}=p(x_t,t)\dot x_t+\int_{-\infty}^{x_t}\partial_tp(y,t)dy=0.$$ Thus, demanding that $x_t=x_0=$ const. for all $t\geq 0$ implies $\int_{-\infty}^{x_0}\partial_tp(y,t)dy=0$. Using the FPE and integrating by parts, we get $$\big(g(x_0)g'(x_0)-f(x_0)\big)F(t)+\frac12g^2(x_0)G(t)=0,$$ where $F(t)=p(x_0,t)$ and $G(t)=\partial_xp(x_0,t)$.

From this, a sufficient condition for $x_0$ to be a fixed $P_{\ast}$-percentile is that both $f$, and $g$ have a root at $x_0$. Intuitively, this means that $x_0$ is a boundary that cannot be crossed by the particle following the diffusion process described by $f$ and $g$, and therefore the fraction $P_{\ast}$ of the total probability mass does not change.

If $F(t)$ and $G(t)$ are linearly independent, this condition is also necessary. One may now ask under what conditions on $f$ and $g$ we obtain $G(t)=\alpha F(t)$ for some $\alpha\in\mathbb{R}$ and all $t\geq 0$, in which case there could be solutions with $f$ or $g$ non-zero at $x_0$.