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[From Bass R.F. Stochastic processes. Exercise 10.4]

Let $N_t = \int_0^tH_sdM_s$ where $M$ is a continuous square integrable martingale and H is predictable and integrable and $L_t = \int_0^tK_sdN_s$ where $K$ is predictable and integrable. Show that

$$ L_t = \int_0^tH_sK_sdM_s $$

As $N_t = \int_0^tH_SdM_s$, differentiating we get: $$ dN_t = H_SdM_s $$ and substituting into $L_t$ we get the required result.

Here I feel as if I am abusing mathematical notation and fear I might go to hell for it - is there another way of about it? Thanks.

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Yes, this is abuse of notation. One way to prove this statement rigorously is to use the characterization of the stochastic integral by its (predictable) covariation:

Theorem Let $X$ a square integrable martingale and $f$ predictable such that $$\mathbb{E} \left( \int_0^{\infty} f^2 \, d\langle X \rangle_s \right) < \infty.$$ Then for any square integrable martingale $Y$, we have $$\langle \int_0^{\cdot} f \, dX, Y \rangle_t = \int_0^t f(s) \, d\langle X,Y \rangle_s \tag{1}.$$ On the other hand, if for some square integrable martingale $I$ the equality $$\langle I,Y \rangle_t = \int_0^t f(s) \, d\langle X,Y\rangle_s \tag{2}$$ holds for all $Y$, then $$I_t-I_0 = \int_0^t f(s) \, dX_s.$$

This means that, in view of $(2)$, we have to prove that $L$ satisfies

$$\langle L,Y \rangle_t = \int_0^t H_s K_s \, d\langle M,Y\rangle_s$$

for any square integrable martingale $Y$. This follows from $(1)$:

$$\begin{align*} \langle L,Y \rangle_t &\stackrel{(1)}{=} \int_0^t K_s \, d\langle N,Y \rangle_s \stackrel{(1)}{=} \int_0^t K_s H_s \, d\langle M,Y \rangle_s. \end{align*}$$

This finishes the proof. Note that we have to assume that $K$ is suitable integrable; otherwise the integral $\int_0^t K_s dN_s$ might not exist.

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