# Derivative of the most probable value (of a gaussian variable) VS most probable value of the derivative

Let $x$ be a random variable with gaussian probability distribution $P(x)$. We assume that $x$ depends parametrically on a parameter $t$ so that : $P(x(t))=\frac{1}{\sqrt{2\pi\sigma^2(t)}}\exp(-\frac{(x-\mu(t))^2}{2\sigma^2(t)})$. The functions $\mu(t)$ and $\sigma(t)$ are supposed to be known.

Let now define a new random variable $y=\frac{dx}{dt}$. My question is : is the most probable value of $y$ (denoted by $y_0$) given by the derivative of the most probable value of $x$, i.e. do we have $y_0=\frac{d\mu(t)}{dt}$? Numerically, it seems to be the case but I'm not able to prove it.

Two remarks:

1. If $y$ was also a gaussian variable then $y_0$ would coincide with the mean value of $y$ (denoted by $\langle y \rangle$) and we would have immediately : $y_0=\langle y \rangle=\langle \frac{dx}{dt} \rangle=\frac{d\langle x \rangle}{dt}=\frac{d\mu(t)}{dt}$.

2. In the problem I study (a particular physical system), $y$ happens to be a lorentzian variable (numerically checked). So $\langle y \rangle$ is infinite and the above justification does not hold anymore. Yet, the result $y_0=\frac{d\mu(t)}{dt}$ seems to be still satisfied according to my numerical results.

Thanks in advance for any help.

-
I don't quite understand what you mean by $P(x(t))$. Written in this way, $P$ is a functional, so a function of the function $x(t)$. But the expression you give for $P$ is not a functional, but a function of two variables, $x$ and $t$. If indeed that is what you intend, $P(x,t)$ instead of $P[x(t)]$, then I do not know what you mean by $y=dx/dt$. Since $x$ does not itself depend on $t$, there is no way to take the derivative. Apologies for my confusion. – Carlo Beenakker Nov 17 '12 at 14:48
Even in the simple case where $\mu(t)$ and $\sigma(t)$ are constant, the stated hypotheses would allow $x(t)$ to be chosen independently for every single $t$, so there is no reason to think $x(t)$ is a differentiable function of $t$. A more realistic example, though with $\sigma$ no longer constant, would be Brownian motion; again, $dx/dt$ won't exist. – Andreas Blass Nov 17 '12 at 15:11
Thanks a lot for your comments and sorry for the confusing notations. Here $x:t\to x(t)$ is a function from an interval $\lbrack a,b \rbrack$ to the real ensemble and it is assumed to be differentiable so that $\frac{dx}{dt}$ does exist. Then we assume that for any (fixed) $t$, $x(t)$ is a gaussian random variable with a center and a variance depending on $t$, i.e. with probability distribution $P(x(t))=\frac{1}{\sqrt{2\pi\sigma^2(t)}}\exp(-\frac{(x(t)-\mu(t))^2}{2\sigma^2(‌​t)})$. I hope it makes sense like this – gfleury Nov 17 '12 at 17:54
to be able to say something about the distribution of $y=dx/dt$ you would need to know how $x(t)$ and $x(t+\delta t)$ are correlated; the distribution $P$ does not contain that information. – Carlo Beenakker Nov 17 '12 at 18:18
Thanks again for your comment. I understand that we cannot give a general answer to this question and that we need additional hypothesis. But if we know the general form of the distribution of $y$ at fixed arbitrary $t$ (gaussian or lorentzian in my two remarks above or something else), then I wonder if $y_0$ can be related to $\frac{d\mu(t)}{dt}$. For a gaussian distribution of $y$, it is straightforward. For a lorentzian distribution of $y$, it seems to be also the case numerically but if it is indeed so, the proof is less obvious. – gfleury Nov 18 '12 at 16:10