Linear stochastic differential equation solution proof

Question

I'm studying from Daniel Revuz, Marc Yor - Continuous Martingales and Brownian Motion book. In particular, I'm having a hard time understanding a particular step into the proof of the Proposition 2.3 of the IX chapter - Stochastic Differential Equation. Below the text of the proposition.

The solution to the linear equation $$ Y_t = H_t + \int_0^t Y_s \text{d}X_s $$ where $H,X$ are two given continuous semimartingales is $$ Y_t = \mathcal{E}(X)_t \left( H_0 + \int_0^t \mathcal{E}(X)_s^{-1} \left( \text{d}H_s - \text{d}\left<H,X \right>_s \right) \right) $$ where $\mathcal{E}(X)_s = \exp(X_s - 1/2 \left<X \right>_s)$ is the stochastic exponential.

Proof. Let us compute $\int_0^t Y_s \text{d} X_s$ for $Y$ given in the statement: $$ \int_0^t Y_s \text{d}X_s = H_0 \int_0^t \mathcal{E}(X)_s \text{d} X_s + \overbrace{\int_0^t \mathcal{E}(X)_s \left( \int_0^s \mathcal{E}(X)_u^{-1} (\text{d}H_u - \text{d}\left<H,X \right>_u)\right) \text{d}X_s}^{(\star)}. $$ In the book's proof the $(\star)$ addend is written as: $$ \int_0^t \mathcal{E}(X)_s \text{d}X_s \left( \int_0^s \mathcal{E}(X)_u^{-1} (\text{d}H_u - \text{d}\left<H,X \right>_u)\right) \overbrace{=}^{(\ast)} $$ and to that addend the integration by parts is computed.

My question is: how was it possible to "move" the $\text{d}X_s$ and ignore the process $$ \int_0^s \mathcal{E}(X)_u^{-1} (\text{d}H_u - \text{d}\left<H,X \right>_u)? $$ Edit. From $(\ast)$ I get $$ \overbrace{=}^{(\ast)} -H_0 + H_0 \mathcal{E}(X)_t + \mathcal{E}(X)_t \int_0^t \mathcal{E}(X)_s^{-1} (\text{d}H_s - \text{d}\left<H,X \right>_s) - \int_0^t \mathcal{E}(X)_s \left( \mathcal{E}(X)_s^{-1} \text{d}H_s - \mathcal{E}(X)_s^{-1} \text{d}\left<H,X \right>_s \right) - \left<\mathcal{E}(X), \int_0^\cdot \mathcal{E}(X)_s^{-1} \text{d}H_s \right>_t $$

Answer 1

Revuz & Yor write $$ \int_0^t\mathcal{E}(X)_s\,\color{red}{dX_s} \left( \int_0^s \mathcal{E}(X)_u^{-1} (\,dH_u - \,d\left<H,X \right>_u)\right) $$ for $$ \int_0^t\mathcal{E}(X)_s \left( \int_0^s \mathcal{E}(X)_u^{-1} (\,dH_u - \,d\left<H,X \right>_u)\right) \,\color{red}{dX_s}\ $$ which is what sometimes physicists do in multiple integrals with complicated integrand. We shall see below that it also is a smart way of grouping together what belongs together:

With the abbreviations $M_t:=\mathcal{E}(X)_t$ and $$ N_t:=\int_0^t \mathcal{E}(X)_u^{-1} (\,dH_u - \,d\left<H,X \right>_u) $$ the integration by parts formula $$ M_tN_t=\int_0^tN_s\,dM_s+\int_0^tM_s\,dN_s+\langle M,N\rangle_t $$ gives \begin{align} &\int_0^t\underbrace{\color{red}{\mathcal{E}(X)_s\,dX_s}}_{dM_s}\underbrace{\left( \int_0^s \mathcal{E}(X)_u^{-1} (\,dH_u - \,d\left<H,X \right>_u)\right)}_{N_s}\\ &=\underbrace{\mathcal{E}(X)_t}_{M_t}\;\underbrace{\int_0^t \mathcal{E}(X)_u^{-1} (\,dH_u - \,d\left<H,X \right>_u)}_{N_t}\\ &-\int_0^t\underbrace{\mathcal{E}(X)_s}_{M_s}\;\underbrace{\left(\mathcal{E}(X)_s^{-1}\,dH_s - \mathcal{E}(X)_s^{-1}\,d\left<H,X \right>_s\right)}_{dN_s}\\ &-\left\langle \underbrace{\mathcal{E}(X)}_{M},\underbrace{\int_0^. \mathcal{E}(X)_u^{-1} (\,dH_u - \,d\left<H,X \right>_u)}_{N}\right\rangle_t \end{align} which is exactly what they wrote.

After having this book in my shelf now for 35 years or so I still find gems in it.