I wanted to describe the time-evolution of a probability density function (PDF) which converge to a dirac distribition centered at $\mu$, denoted as $\delta_\mu(x)$. To achieve that, I used the inverse function of the cumulative distribution function (CDF), the quantile function, denoted $Q(p)$. I tried the differential eq:
$${\partial Q\over\partial t}=\mu-Q $$
After solve it:
$$Q=\mu+e^{-t}(Q_0-\mu)$$
Then, applying the fact that $\Bbb E[X^k]=\int_{[0,1]}Q^k dp $ $$\Bbb E[X_t^k]=\int_0^1 Q^k dp=\int_0^1 \left(\mu+e^{-t}(Q_0-\mu)\right)^k\ dp $$ then by dominated convergence theorem, I can take the limit inside the integral, given that $Q\to\mu$ as $t\to\infty$ for all $p$...
$$\lim_{t\to\infty}\Bbb E[X_t^k]=\int_0^1 \lim_{t\to\infty} \left(\mu+e^{-t}(Q_0-\mu)\right)^k\ dp $$
$$=\int_0^1 \mu ^k\ dp=\mu ^k $$
Now, to work back with the CDF, we can invert Q, or use the fact that:
$${\partial Q\over \partial t}=\psi(x)\phi(t)\Rightarrow {\partial F\over \partial t}=-\psi(x)\phi(t){\partial F\over \partial x} $$
where $\psi$ and $\phi$ are function of $x$ and $t$, and $F$ is the CDF of r.v X. So the PDE which describe the time-evolution of any CDF to the CDF of the dirac distribution is:
$${\partial F\over \partial t}=(x-\mu){\partial F\over \partial x} $$ and for the time-evolution of the PDF, we differentiate with respect $x$.
$${\partial \over\partial x}{\partial F\over \partial t}={\partial \over \partial x} \left((x-\mu){\partial F\over \partial x}\right)$$
$${\partial f\over \partial t}=f+(x-\mu){\partial f\over \partial x} $$
where $f$ is the PDF of r.v X.
The solution of that PDE is:
$$e^{t}f_0(\mu+e^{t}(x-\mu))$$
where $f_0$ is the initial distribution. Now, my question is how you can describe the same convergence of all distribution to a dirac distribution, using the lenguage of SDE?
My attempt is to describe the PDE $\partial_t Q=-Q$ as
$$dX_t = -X_t dt $$
because $Q(p)=x$