[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.


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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