# How does the general stochastic integral generalize the Brownian motion stochastic integral?

I've been reading in the book "Continuous martingales and Brownian motion" about the definition of a general stochastic integral with respect to a bounded continuous martingale. The definition goes as follows:

Given a continuous bounded second order martingale $M$, and a progressively measurable process $K$ satisfying $\mathbb{E}\Big[ \int_0^t K_s^2d\langle M,M\rangle_s \Big]<\infty$.

we define the stochastic integral of $K$ with respect to $M$ as the unique continuous bounded martingale vanishing at $0$ ,$U$, such that:

$\langle U, N \rangle_t= \int_0^t K_s d\langle M,N \rangle_s$ for all continuous bounded second order martingale $N$.

What I am unsure on, is how does this property generalize the standard Brownian motion stochastic integral, i.e how does one prove that:

$\Big \langle \int_0^t K_sdB_s , N \Big \rangle_t= \int_0^t K_s d\langle B,N \rangle_s$ for all continuous bounded second order martingale $N$, where $B$ is a brownian motion.

I would appreciate any hints or answers on the matter (though it may be quite simple, but I cannot see the answer right now).

No, it's not exactly obvious. Let's prove the following theorem.

Let $(B_t)_{t \geq 0}$ be a one-dimensional Brownian motion and $(K_t)_{t \geq 0}$ a progressively measurable process such that $\mathbb{E}\int_0^t K_s^2 \, ds < \infty$ for all $t \geq 0$. If $(N_t)_{t \geq 0}$ is a continuous $L^2$-bounded martingale, then $$M_t := \int_0^t K_s \, dB_s$$ satisfies $$\langle M,N \rangle_t = \int_0^t K_s \, d\langle B,N \rangle_s.$$

Proof: Througout, $(N_t)_{t \geq 0}$ is an $L^2$-bounded martingale. First we consider the particular case that $K$ is a simple process of the form $$K(s) = \sum_{j=0}^{N-1} \varphi_j 1_{(s_j,s_{j+1}]}(s) \tag{1}$$ where $0<s_0 < \ldots < s_N$ and $\varphi_j \in L^2(\mathcal{F}_{s_j})$. We have to show that $M_t = \int_0^t K_r \, dB_r$ satisfies $$\mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) = \mathbb{E} \left( \int_s^t K_r \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right) \tag{2}$$ for any fixed $s \leq t$. Without loss of generality, we may assume that $s_N = t$ and that there exists $k \in \{0,\ldots,N\}$ such that $s_k = s$ (otherwise we refine the partition accordingly). Writing $$N_t-N_s = \sum_{i=k}^{N-1} (N_{s_{i+1}}-N_{s_i}) \quad \text{and} \quad M_t-M_s = \sum_{j=k}^{N-1} (M_{s_{j+1}}-M_{s_j})$$ we find

$$\mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) = \sum_{j=k}^{N-1} \sum_{i=k}^{N-1} \mathbb{E}((M_{s_{j+1}}-M_{s_j}) (N_{s_{i+1}}-N_{s_i}) \mid \mathcal{F}_s).$$

Since both $(M_t)_{t \geq 0}$ and $(N_t)_{t \geq 0}$ are martingales, it is not difficult to see from the tower property that the terms on the right-hand side vanish for $i \neq j$, and so $$\mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) = \sum_{j=k}^{N-1} \mathbb{E}(\varphi_j (B_{s_{j+1}}-B_{s_j}) (N_{s_{j+1}}-N_{s_j}) \mid \mathcal{F}_s).$$

Using once more the tower property, we get

\begin{align*} \mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) &= \sum_{j=k}^{N-1} \mathbb{E} \bigg[ \varphi_j \mathbb{E}((B_{s_{j+1}}-B_{s_j}) (N_{s_{j+1}}-N_{s_j}) \mid \mathcal{F}_{s_j}) \mid \mathcal{F}_s \bigg] \\ &= \sum_{j=k}^{N-1} \mathbb{E} \bigg[ \varphi_j \mathbb{E}(\langle B,N \rangle_{s_{j+1}}-\langle B,N \rangle_{s_j}) \mid \mathcal{F}_{s_j}) \mid \mathcal{F}_s \bigg]\\ &= \mathbb{E} \left( \sum_{j=0}^{N-1} \varphi_j (\langle B,N \rangle_{s_{j+1}}-\langle B,N \rangle_{s_j}) \mid \mathcal{F}_s \right) \\ &= \mathbb{E} \left( \int_s^t K_r \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right) \end{align*}

This proves the assertion for the simple process $K$. For the general case we choose a sequence of simple process $(K_n)_{n \in \mathbb{N}}$ of the form $(1)$ such that $$\mathbb{E} \left( \int_0^t (K_n(s)-K(s))^2 \, ds \right) \to 0.$$ By the construction of the stochastic integral, this implies, in particular,

$$\int_0^t K_n(r) \, dB_r \to \int_0^t K(r) \, dB_r \quad \text{in L^2(\mathbb{P})} \tag{3}$$

On the other hand, it follows from the Cauchy-Schwarz inequality that

$$\int_0^t K_n(r) \, d\langle B,N \rangle_r \to \int_0^t K(r) \, d\langle B,N \rangle_r \quad \text{in L^2(\mathbb{P})} \tag{4}$$

Thus, $M_t = \int_0^t K(r) \, dB_r$ satisfies

\begin{align*} \mathbb{E} \left( (M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s \right) &\stackrel{(3)}{=} \lim_{n \to \infty} \mathbb{E} \left[ \left( \int_0^t K_n(r) \, dB_r - \int_0^s K_n(r) \, dB_r \right) (N_t-N_s) \mid \mathcal{F}_s \right] \\ &\stackrel{(1)}{=} \lim_{n \to \infty} \mathbb{E} \left( \int_s^t K_n(r) \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right) \\ &\stackrel{(4)}{=} \mathbb{E} \left( \int_s^t K(r) \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right). \end{align*}

• Thank you very much for a very thorough answer. I assumed there could be a more elegant (without using approximation by simple proceeses) for this. – Keen-ameteur Aug 7 '18 at 18:29
• @Keen-ameteur I'm not aware of a proof which doesn't use an approximation argument ... obviously this doesn't mean that such a proof doesn't exist. – saz Aug 8 '18 at 6:12
• @saz I also had a problem seeing this equality in Yor's book. A part of the proof establishes that $(K \cdot M )N -K<M,N>$ is a martingale, and this fact I cannot put into corresponande with any part of the statement. Could one use that to prove the above equality? – user1 Oct 4 '18 at 5:03
• @saz ah I got it! I follows since the qudratic variation process is the unique process for which this is a martingale – user1 Oct 4 '18 at 11:10
• @user1 Yes, but you have to be a bit careful about measurability. In general, there might be several processes $X$ such that $(K \cdot M)N-X$ is a martingale; but only one has is adapted to the right $\sigma$-algebra. – saz Oct 5 '18 at 4:46