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

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

  • $\begingroup$ 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. $\endgroup$ – Keen-ameteur Aug 7 '18 at 18:29
  • $\begingroup$ @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. $\endgroup$ – saz Aug 8 '18 at 6:12
  • $\begingroup$ @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? $\endgroup$ – user1 Oct 4 '18 at 5:03
  • $\begingroup$ @saz ah I got it! I follows since the qudratic variation process is the unique process for which this is a martingale $\endgroup$ – user1 Oct 4 '18 at 11:10
  • $\begingroup$ @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. $\endgroup$ – saz Oct 5 '18 at 4:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.