这是一份nottingham诺丁汉大学ECON1046作业代写的成功案例

Let $\lambda_{1}$ be in $(0, \delta)$. Then, resorting to Taylor’s Theorem, ${ }^{5}$ there exists a $\theta \in(0,1)$ such that
$$
g_{x(t)}\left(\lambda_{1}\right)=\frac{\lambda_{1}^{2}}{2} g_{x(t)}^{\prime \prime}\left(\theta \lambda_{1}\right)>0=g_{x(t)}(0) .
$$
This, however, contradicts our earlier observation that $g_{x}(\lambda)$ attains its maximum at $\lambda=0$ for all $\boldsymbol{x}{0}$ and $\boldsymbol{x}$ of $E$. (6) (Sufficiency). Let $\boldsymbol{x}{0}, \boldsymbol{x}$ and $g_{\boldsymbol{x}}(\lambda)$ be as in the proof of necessity.
Then, by hypothesis,
$$
g_{x}^{\prime \prime}(\lambda) \leqq 0 \text { for all } x_{0}, x \in E \text {, and for all } \lambda \in[0,1] .
$$
This, together with the fact that $g_{x}^{\prime}(0)=0$ for any $\boldsymbol{x}{0}$ and $x$ of $E$, implies that $$ g{x}^{\prime}(\lambda) \leqq 0 \text { for all } x_{0}, x \in E \text {, and for all } \lambda \in[0,1] .
$$
Since, by definition, $g_{x}(0)=0$ for all $x_{0}$ and $x$ of $E$, we can similarly assert that $g_{x}(\lambda) \leqq 0$ for all $x_{0}, x \in E$ and for all $\lambda \in[0,1] .$

ECON1046 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
$$
\begin{aligned}
z_{t} &=\tau_{y t}+c_{y t}+\epsilon_{y t}-\lambda\left(\alpha \tau_{y t}+c_{x t}+\epsilon_{x t}\right) \
&=\tau_{y t}-\lambda \alpha \tau_{y t}+c_{y t}-\lambda c_{x t}+\epsilon_{y t}-\lambda \epsilon_{x t}
\end{aligned}
$$