这是一份imperial帝国理工大学 M45A51作业代写的成功案例
$$
d q_{t}=\beta q_{t} d t+\theta \gamma(\theta) d W_{t}, q_{0}=0 .
$$
It is easy to verify that
$$
q_{t}=\theta \gamma(\theta) \int_{0}^{t} e^{-\sqrt{\theta^{2}+\beta^{2}(t-s)}} d X_{s} .
$$
In this case the LAMN property holds with
$$
m_{T}(\theta)=e^{\theta T}, \quad \zeta(\theta)=\left(\int_{0}^{\infty} e^{-\sqrt{\theta^{2}+\beta^{2}}(t-s)} d W_{s}\right)^{2} .
$$
M45A51 COURSE NOTES ：
The indefinite integral is defined as
$$
\int_{0}^{t} \psi(s) d W_{s}^{H}=\int_{0}^{1} \psi(t) I_{[0, t]} d W_{t}^{H}
$$
This integral has a continuous version and a Gaussian process. However,
$$
E\left(\int_{0}^{t} \psi(s) d W_{s}^{H}\right) \neq 0 .
$$
To overcome this situation, Duncan, Hu and Pasik-Duncan (2000) introduced an integral using Wick calculus for which
$$
E\left(\int_{0}^{t} f(s) d W_{s}^{H}\right)=0 .
$$