# How to obtain this differential relation about moments of a stochastic process?

$$\newcommand{\Ex}{\mathbb E}$$ 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.

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 $$\mu_{2 n}^{\prime}(t)=n\left\{2(2 n-1) \mu_{2 n-2}(t) - \Ex \left[Y_t^{2 n-1}\left(b\left(t, X_t\right)-b\left(t, X_t^{\prime}\right)\right)\right]\right\}. \tag{1}\label{1}$$

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 $$\ref{1}$$?

I believe you meant lemma 3.8. Here at the final step of your calculation, you just apply 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.