这是一份nottingham诺丁汉大学ECON2004作业代写的成功案例

$$
\sigma_{t}^{2}=\sigma^{2} \exp g(t), \quad \text { or } \quad \log \sigma_{t}^{2}=\log \sigma^{2}+g(t)
$$
where $g(t)$ is a deterministic function that can represent a diurnal pattern and starts at zero, that is $g(0)=0$. The function $g(t)$ is typically very smooth so that deviations from a diurnal pattern are not captured by $g(t)$. An example of an appropriate specification for $g(t)$ is given in Appendix A. The integrated volatility becomes
$$
\sigma^{* 2}(0, t)=\int_{0}^{t} \sigma_{s}^{2} \mathrm{~d} s=\sigma^{2} \int_{0}^{t} \exp g(s) \mathrm{d} s
$$
The actual volatility can be analytically derived from (8) or it can be approximated by
$$
\sigma^{* 2}\left(t_{n}, t_{n+1}\right) \approx \sigma^{2} \sum_{s=l_{n}}^{t_{s+1}} \exp g(s)
$$

ECON2004 COURSE NOTES ：

A consistent estimator of $\Omega_{}$ can easily be obtained by a Bartlett kernel estimator, i.e.: $$ \hat{\Omega}{\text {予 }}=\hat{\Gamma}{0}+\sum_{k=1}^{K(T)}\left(1-\frac{k}{K(T)+1}\right)\left(\hat{\Gamma}{k}+\hat{\Gamma}{k}^{\prime}\right)
$$
where
$$
\hat{\Gamma}{k}=\frac{1}{T} \sum{t=k+1}^{T}\left[g_{t-k}(\hat{u})-\mu(\theta)\right]\left[g_{t}(\hat{u})-\mu(\theta)\right]^{\prime}
$$
with $\theta$ replaced by a consistent estimator $\tilde{\theta}{T}$ of $\theta$. The truncation parameter $K(T)=\tilde{c} T^{1 / 3}$ is allowed to grow with the sample size such that: $$ \lim {T \rightarrow \infty} \frac{K(T)}{T^{1 / 2}}=0
$$
A consistent estimator of $V_{}\left(\theta_{0}\right)$ is then given by
$$
\hat{V}{}=\left[P\left(\hat{\theta}{T}\right) \hat{\Omega}{}^{-1} P\left(\hat{\theta}{T}\right)^{\prime}\right]^{-1} .
$$