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:

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 : $y_0=\langle y \rangle=\langle \frac{dx}{dt} \rangle=\frac{d\langle x \rangle}{dt}=\frac{d\mu(t)}{dt}$.

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.