$\newcommand{\Ex}{\mathbb E}$ 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 $\ref{1}$?