这是一份umass麻省大学 MATH 437作业代写的成功案例
To see this, note that the increments of $M_{t}$ over small intervals $\Delta$ will be given by
$$
\Delta M_{t}=\Delta N_{t}^{C}-\Delta N_{t}^{B}
$$
Apply the conditional expectation operator:
$$
E_{l}\left[\Delta M_{t}\right]=E_{t}\left[\Delta N_{t}^{G}\right]-E_{t}\left[\Delta N_{f}^{B}\right]
$$
But, approximately,
$$
\begin{aligned}
E_{f}\left[\Delta N_{t}^{G}\right] & \cong 0 \cdot(1-\lambda \Delta)+1 \cdot \lambda \Delta \
& \cong \lambda \Delta
\end{aligned}
$$
and similarly for $E_{t}\left[\Delta N_{t}^{B}\right]$. This means that
$$
E_{t}\left[\Delta M_{i}\right] \cong \lambda \Delta-\lambda \Delta=0 .
$$
MMATH 437 COURSE NOTES ：
The discrete equivalent of the martingale representation in is then given by the following equation:
$$
C_{T}=C_{t}+\sum_{i=1}^{n} D_{t_{i}} \Delta+\sum_{i=1}^{n} g\left(C_{t_{t}}\right) \Delta M_{t_{i}}
$$
where $\Delta M_{t_{\mathrm{r}}}$ means
$$
\Delta M_{t_{1}}=M_{t_{i-1}}-M_{t_{j}}
$$
and $n$ is such that
$$
t_{v}=t<\ldots<t_{n}=T .
$$