In a 1966 paper by the physicist Kubo (https://iopscience.iop.org/article/10.1088/0034-4885/29/1/306), I found the following problem: The author considers an OU process of the form
$$m\frac{dv}{dt}=-m\gamma v +R(t)$$
Where the second term is a gaussian stochastic process. He then argues that the probability distribution of velocities of this object should be given by
\begin{align} \frac{\partial P}{\partial t}(t,v)=\nabla_v\cdot\left[D_v\nabla_v +\gamma v\right]P && (1.1) \end{align}
Then he argues that these quantities should be related by the expression
$$ D_v=\frac{1}{m^2}\int_0^\infty \langle R(0)R(s)\rangle ds$$
However, if I were to write this SDE using standard Ito calculus notation, I would write
$$dv=-\gamma v dt +\sqrt{2D_v}dW_t$$
In order to reach the same PDE as in equation (1.1). However, if I do that, that would mean that $R(s)=m\sqrt{2D_v}dW_t$, which would mean
$$\frac{1}{m^2}\int_0^\infty \langle R(0)R(s)\rangle ds=2D_v\int_0^\infty \langle dW_0 dW_s\rangle ds=2D_v\int_0^\infty \delta(s) ds=2D_v$$
Why did I get this extra factor of 2? Where did I go wrong? Is he using a nonstandard notation?
Must be because the heat equation $$ \partial_tp=\partial_x^2p $$ is satisfied by the heat kernel $$ p_t(x,y)=\frac{1}{\sqrt{\color{red}{4}\,\pi\,t}}e^{-\frac{(x-y)^2}{\color{red}{4}\,t}} $$ whilst standard Brownian motion has transition probability $$ p_t(x,y)=\frac{1}{\sqrt{\color{red}{2}\,\pi\,t}}e^{-\frac{(x-y)^2}{\color{red}{2}\,t}} $$ that satisfies $$ \partial_tp=\color{red}{\frac12}\partial_x^2p\,. $$