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

Question

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

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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 $\ref{1}$?

Answer 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.