$\newcommand{\Ex}{\mathbb E}$ I'm reading an argument in the proof of *Proposition 3.8.* in the paper [Nonlinear self-stabilizing processes - I Existence, invariant probability, propagation of chaos](https://www.sciencedirect.com/science/article/pii/S0304414998000180).




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Let $X\_0$ and $X\_0^{\prime}$ be two independent real-valued random variables, having the same distribution. We consider two independent $1$-dimensional Brownian motions $B$ and $B'$. Let $X$ and $X'$ be solutions of
$$
X\_t=X\_0+B\_t-\frac{1}{2} \int\_0^t b\left(s, X\_s\right) \mathrm{d} s,
$$
and
$$
X\_t^{\prime}=X\_0^{\prime}+B\_t^{\prime}-\frac{1}{2} \int\_0^t b\left(s, X\_s^{\prime}\right) \mathrm{d} s,
$$


where $b:\mathbb R\_{\ge 0} \times \mathbb R \to \mathbb R$ is regular enough. Let
$$
Y\_t := X\_t-X\_t^{\prime}
\quad \text{and} \quad
\mu\_n(t) := \Ex \left ( \left|Y\_t\right|^n \right), \quad n \geqslant 2 .
$$


Then $Y$ is a semi-martingale with decomposition
$$
Y\_t=Y\_0+B\_t-B\_t^{\prime}-\frac{1}{2} \int\_0^t\left(b\left(s, X\_s\right)-b\left(s, X\_s^{\prime}\right)\right) \mathrm{d} s .
$$


We apply the Itô's formula and take the expectation and the derivative. We obtain
$$
\mu\_{2 n}^{\prime}(t)=n\left\{2(2 n-1) \mu\_{2 n-2}(t) - \Ex \left[Y\_t^{2 n-1}\left(b\left(t, X\_t\right)-b\left(t, X\_t^{\prime}\right)\right)\right]\right\}.
\tag{1}\label{1} 
$$




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**My understanding** By Itô's lemma,
$$
\begin{align\*}
& (Y\_t)^{2n} - (Y\_0)^{2n} \\
= & 2n \int\_0^t (Y\_s)^{2n-1} \mathrm{d} Y\_s + \frac{2n(2n-1)}{2} \int\_0^t (Y\_s)^{2n-2} \mathrm{d} \langle Y \rangle \_s.
\end{align\*}
$$


We have
$$
\begin{align\*}
\mathrm{d} Y\_s &= \mathrm{d} B\_s - \mathrm{d} B'\_s- \frac{\left(b\left(s, X\_s\right)-b\left(s, X\_s^{\prime}\right)\right) }{2} \, \mathrm{d} s, \\
\mathrm{d} \langle Y \rangle\_s &= \mathrm{d} \langle B - B' \rangle \_s = 2 \, \mathrm{d} s.
\end{align\*}
$$


Hence
$$
\begin{align\*}
\Ex[(Y\_t)^{2n}] - \Ex [(Y\_0)^{2n}] &= -n \int\_0^t \Ex [ (Y\_s)^{2n-1} \left(b\left(s, X\_s\right)-b\left(s, X\_s^{\prime}\right)\right)] \, \mathrm{d} s \\
& \qquad + 2n(2n-1) \int\_0^t \Ex [ (Y\_s)^{2n-2} ] \, \mathrm{d} s.
\end{align\*}
$$


So
$$
\begin{align\*}
\mu\_{2n} (t) - \mu\_{2n}(0) &= -n \int\_0^t \Ex [ (Y\_s)^{2n-1} \left(b\left(s, X\_s\right)-b\left(s, X\_s^{\prime}\right)\right)] \, \mathrm{d} s \\
& \qquad + 2n(2n-1) \int\_0^t \mu\_{2n-2} (s) \,\mathrm{d} s.
\tag{2}\label{2} 
\end{align\*}
$$



> 
> Could you please explain how to go from $\ref{2}$ to $\ref{1}$?
> 
> 
> 



How to obtain this differential relation about moments of a stochastic process?
I believe you meant lemma 3.8. Here at the final step of your calculation, you just apply [Lebesgue-differentiation theorem](https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem) since those quantities are continuous and the integral is Lebesgue. That will give you the $\mu'(0)$.


So to get the $\mu'(t\_{0})$, I would apply Ito's formula with starting points $t=t\_{0}$ as opposed to $t=0$. So the integrals will be $\int\_{t\_{0}}^{t\_{0}+t}$. Meaning apply Ito's lemma separately to $t\_{0}$ and to $t\_{0}+t$ and then take their difference.


