Cross Variation of Ito Integral with $L^2$ bounded continuous martingale

Question

I am trying to prove the identity:

$$ < I(K), N_{\cdot}>_t = \int_0^t K_s \ d < B,N_{\cdot}>_s $$

where $I(K)_t$ is by definition $ \int_0^t K_s \ d B_s $, and we take $K$ as a progressively measurable process, bounded with respect to the norm $\mathbb E[ \int_0^{\infty} (K_s)^2 \ ds]$, and we take $N$ to be a $L^2$ bounded martingale ( $\sup_{s \geq 0} \mathbb E[ (N_s)^2 ] < \infty $).

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My attempt:

The problem I am facing is that I do not know how to rewrite $d < B,N>_s$ or simplify the following term:

\begin{align} < I(K), N_{\cdot}>_t &= < \int_0^{\cdot} K_s \ d B_s, N_{\cdot}>_t \\ &= \frac 1 4 [ < \int_0^{\cdot} K_s \ d B_s + N_{\cdot}>_t - < \int_0^{\cdot} K_s \ d B_s - N_{\cdot}>_t] \\ &= \ ?? \\ &= \int_0^t K_s \ d < B,N_{\cdot}>_s \end{align}

I thought about using the polarising identity: $<M-N> = <M> + <N> - 2 <M,N>$ but I am unsure about how to use it correctly in this case.

Any thoughts?

Answer 1

After working it and using hints, here I described the steps to do the proof:

We prove for all $M,N$ local martingales: $< \int H d M, N > = \int H \ d < M,N> $

1. Take $M$ and $N$ $L^2$ continuous martingales first. $H$ a simple process. Then using the orthogonality of martingales:

\begin{align} < \int H \ d M, N >_t &= < \sum H_{t_i} ( M^{t_{i+1}} - M^{t_{i}}), N >_t \\ &= \sum H_{t_i} ( <M,N>_t^{t_{i+1}} - <M,N>_t^{t_{i}}) \\ &= \int H \ d < M,N> \end{align}

2. Replace $H$ by a progressively bounded process ( bounded in the sense: $\mathbb E[ \int^{\infty}_0 H_s^2 \ d < M >_s ]$, and use the Kunita Watanabe inequality.

3. Conclude with stopping times (localisation argument) and replacing $M,N$ local martingales.

There is a very nice proof of this statement (thrm 5.4) in Le Gall Brownian Motion, Martingales, and Stochastic Calculus.