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How to proof that: $$\langle X\cdot M , Y\cdot N\rangle_T = \int^T_0 XY d \langle M, N\rangle$$ for $X,Y$ - simple processes and $M,N \in M^2_c$.

If I need to prove that $$\langle X\cdot M \rangle_T = \int^T_0X^2 d \langle M\rangle$$ then knowing that DM decomposition tells us that: $X^2 - \langle X\rangle$ is a martingale, I can take conditional expectation and show that: $$ \mathbb{E}\left[(X \cdot M)_T^2 - \langle X\cdot M \rangle_T - (X \cdot M)_S^2 + \langle X\cdot M \rangle_S | \mathcal{F}_S\right] = 0$$ What is the way to deal with $\langle X\cdot M , Y\cdot N\rangle_T$?

I suspect that working the same way I will get something similar, but I can not continue: $$ \mathbb{E}\left[(X \cdot M)_T(Y \cdot N)_T - \langle X\cdot M , Y\cdot N \rangle_T - (X \cdot M)_S(Y \cdot N)_S + \langle X\cdot M, Y\cdot N \rangle_S | \mathcal{F}_S\right] = 0$$ rewriting terms as martingale transforms. Is this a case when we consider simpler case and look at when $X=Y, M=N$?

saz
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StochasticIntegrationStudent
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1 Answers1

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Hint:

  1. Show (or recall) that the quadratic covariation satisfies $$\langle M,N \rangle_t = \frac{1}{4} \big( \langle M+N \rangle_t - \langle M-N \rangle_t \big) \tag{1}$$ for any two martingales $M,N \in \mathcal{M}_c^2$.
  2. Use $(1)$ and the fact that $$\langle X \bullet M \rangle_t = \int_0^t X(s)^2 \, d\langle M \rangle_s$$ to show that $$\langle X \bullet M, Y \bullet M \rangle_t = \int_0^t X(s) Y(s) \, d\langle M \rangle_s. \tag{2} $$
  3. Use $(2)$, $(1)$ and $$\begin{align*} \langle X \bullet M, Y \bullet N \rangle_t &= \frac{1}{4} \bigg( \langle X \bullet (M+N), Y \bullet (M+N) \rangle_t \\ &\qquad - \langle X \bullet (M-N), Y \bullet (M-N) \rangle_t\bigg) \end{align*}$$ to conclude $$\langle X \bullet M, Y \bullet N \rangle_t = \int_0^t X(s) Y(s) \, d\langle M,N\rangle_s.$$

Alternative approach: Since $X$ and $Y$ are simple processes, the stochastic integrals $X \bullet M$ and $Y \bullet N$ can be calculcated explicitly. Then it is a straight-forward computation to verify that $$(X \bullet M)_t \cdot (Y \bullet N)_t - \int_0^t X(s) Y(s) \, d\langle M,N\rangle_s$$ is a martingale, and this proves the assertion.

saz
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  • Dear @saz, you were faster then me ;) wanted to fix typo with $N$. Thank you very much! It is a very elegant way to prove. Could you also give a hint, (maybe smaller) but how to prove in alternative approach? As I see it, I need to write down $X\bullet M$ as a sum, hence: $$(X\bullet M)t(Y\bullet N)_t = \sum^\infty \xi_i( M{t\wedge t_{i+1}}- M_{t\wedge t_{i}} ) \sum^\infty \xi^*i( N{t\wedge t_{i+1}}- N_{t\wedge t_{i}} )$$ but it gets cumbersome. Is that a correct way? – StochasticIntegrationStudent Mar 08 '17 at 18:57
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    @StochasticIntegrationStudent Essentially, the idea is to use that $$\mathbb{E}((M_t-M_r)(N_t-N_r) \mid \mathcal{F}r) = \mathbb{E}(\langle M,N \rangle_t - \langle M,N \rangle_r \mid \mathcal{F}_r) \tag{3}$$ for all $r \leq t$. If you condition the product of the integrals on, say, $\mathcal{F}_s$, then you can pull outside all the terms with $t{i+1} \leq s$; for the other terms use the tower property to condition on $\mathcal{F}{t_i}$ and use $(3)$ (with $r=t_i$) and the fact that $\xi_i$ is $\mathcal{F}{t_i}$-measurable. – saz Mar 08 '17 at 19:04