量化经济学 Quantitative Economics ECON1045

0

这是一份nottingham诺丁汉大学ECON1045作业代写的成功案例

量化经济学 Quantitative Economics ECON1045


Finally, the maximized likelihood function is found from
$$
L_{\max }^{-2 T}=|\hat{Q}|=\left|S_{o o}\right| \prod_{i=1}^{r}\left(1-\hat{\lambda}{i}\right) $$ and the likelihood ratio test of the hypothesis $H{1}(r)$ is given by the trace test statistic
$$
-2 \log Q\left[H_{1}(r) \mid H_{o}\right]=-T \sum_{i=r+1}^{p} \log \left(1-\hat{\lambda}{i}\right) $$ An alternative test (called the maximum eigenvalue test, $\lambda{\max }$ ) is based on the comparison of $H_{1}(r-1)$ against $H_{1}(r)$ :
$$
-2 \log Q\left[H_{1}(r-1) \mid H_{1}(r)\right]=-T \log \left(1-\hat{\lambda}_{r+1}\right)
$$





英国论文代写Viking Essay为您提供作业代写代考服务

ECON1045 COURSE NOTES :

If the individual series have a stochastic trend, we can explore for shared stochastic trends between the series. In particular, if the stochastic trend of $x_{t}$ is shared with the $y_{t}$ series (i.e., $\tau_{x t}$ is linearly related to $\tau_{y t}$ ), then we have the following structure
$$
\begin{aligned}
&y_{t}=\tau_{y t}+c_{y t}+\epsilon_{y t} \
&x_{t}=\alpha \tau_{y t}+c_{x t}+\epsilon_{x t}
\end{aligned}
$$
where $\alpha$ is the factor of proportionality between the two trends. In this case there is a unique coefficient $\lambda$, such that the following linear combination of $y_{t}$ and $x_{t}$
$$
z_{t}=y_{t}-\lambda x_{t}
$$
is a stationary series – see Engle and Granger (1987). In fact, if there is a shared stochastic trend, the linear combination $z_{t}$ can be written as
$$
\begin{aligned}
z_{t} &=\tau_{y t}+c_{y t}+\epsilon_{y t}-\lambda\left(\alpha \tau_{y t}+c_{x t}+\epsilon_{x t}\right) \
&=\tau_{y t}-\lambda \alpha \tau_{y t}+c_{y t}-\lambda c_{x t}+\epsilon_{y t}-\lambda \epsilon_{x t}
\end{aligned}
$$











量化经济学 Quantitative Financial Economics ECON308

0

这是一份liverpool利物浦大学ECON308的成功案例

量化经济学 Quantitative Financial Economics ECON308


$$
F_{j}=S_{i-1, j} \exp (r \Delta t)=K \exp (r \Delta t)
$$
The up jump transition probability, see Equation $10.114$, is
$$
p_{j}=\frac{F_{j}-S_{i, j}}{S_{i, j+1}-S_{i, j}}
$$
which results in a down jump probability of
$$
1-p_{j}=q=1-\frac{F_{j}-S_{i, j}}{S_{i, j+1}-S_{i, j}}=\frac{S_{i, j+1}-F_{j}}{S_{i, j+1}-S_{i, j}}
$$
Multiplying top and bottom by $S_{i, j}$ we obtain
$$
q=\frac{S_{i, j+1}-F_{j}}{S_{i, j+1}-S_{i, j}}=\frac{\left(S_{i, j+1}-F_{j}\right) S_{i, j}}{\left(S_{i, j+1}-S_{i, j}\right) S_{i, j}}=\frac{S_{i, j+1} S_{i, j}-F_{j} S_{i, j}}{S_{i, j+1} S_{i, j}-S_{i, j}^{2}}
$$
We choose to centre at the spot $S_{i, j+1} S_{i, j}=K^{2}$ and we have
$$
q=\frac{S_{i, j+1} S_{i, j}-F_{j} S_{i, j}}{K^{2}-S_{i, j}^{2}}=\frac{S_{i, j+1} S_{i, j}-F_{j} S_{i, j}}{\left(K-S_{i, j}\right)\left(K+S_{i, j}\right)}=\frac{K^{2}-F_{j} S_{i, j}}{\left(K-S_{i, j}\right)\left(K+S_{i, j}\right)}
$$

英国论文代写Viking Essay为您提供作业代写代考服务

ECON308 COURSE NOTES :

Finite-difference approximations for these derivatives can be obtained by considering a Taylor expansion about the point $f_{i, j}$. We proceed as follows:
$$
\begin{aligned}
&f_{i, j+1}=f_{i, j}+f_{i, j}^{\prime} \Delta S+\frac{1}{2} f_{i, j}^{\prime \prime}(\Delta S)^{2} \
&f_{i, j-1}=f_{i, j}-f_{i, j}^{\prime} \Delta S+\frac{1}{2} f_{i, j}^{\prime \prime}(\Delta S)^{2}
\end{aligned}
$$
Subtracting Equations $10.145$ and $10.146$ we obtain:
$$
f_{i, j+1}-f_{i, j-1}=2 f_{i, j}^{\prime} \Delta S
$$
and so
$$
f_{i, j}^{\prime}=\frac{f_{i, j+1}-f_{i, j-1}}{2 \Delta S}
$$