= Solution
For a <simple predictable process>, define its <stochastic integral> by
$$
(H\cdot M)_t=\sum_{k=0}^{n-1}Z_k\bigl(M_{t\wedge t_{k+1}}-M_{t\wedge t_k}\bigr).
$$
It is constant after $t_n$, so the symbol $(H\cdot M)_\infty$ means its value there. Put $\Delta_kM=M_{t_{k+1}}-M_{t_k}$. If $k<\ell$, then $Z_k\Delta_kM Z_\ell$ is $\mathcal F_{t_\ell}$-measurable, and the <conditional expectation> of $\Delta_\ell M$ given that <sigma-algebra> is zero. Boundedness of the coefficients and square-integrability of $M$ justify the expectations, so all cross terms vanish:
$$
\mathbb E[(H\cdot M)_\infty^2]
=\sum_k\mathbb E[Z_k^2(\Delta_kM)^2].
$$
For each diagonal term, the <martingale> increment identity also gives
$$
\mathbb E[Z_k^2(\Delta_kM)^2]
=\mathbb E[Z_k^2(M_{t_{k+1}}^2-M_{t_k}^2)].
$$
Indeed, the omitted term $2Z_k^2M_{t_k}\Delta_kM$ has expectation zero. Apply the assumed <martingale> property of $M^2-[M]$ to replace this difference of squares by the corresponding <quadratic variation> increment. Thus
$$
\boxed{\mathbb E[(H\cdot M)_\infty^2]
=\sum_k\mathbb E[Z_k^2([M]_{t_{k+1}}-[M]_{t_k})]
=\mathbb E[(H^2\cdot[M])_\infty].}
$$
The intervals $(t_k,t_{k+1}]$ in the Lebesgue–Stieltjes integral exactly match these increments, including possible jumps of $M$.
Back to article page
