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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

The goal of a steady-state analysis is to find possible steady states, which can be

• A stationary trajectory (unknown variables are constant in time) or
• A balanced growth path (all variables grow at the same constant rate).
The steady-state analysis plays an important role in economics and is mathematically simpler than a complete dynamic analysis.

Let us find and analyze possible balanced growth paths in the model $(2.31)-(2.35)$. It is easy to see that the original variables $Q(t), C(t), I(t)$, and $K(t)$

of the model grow with the same rate only if the capital-labor ratio $k(t)$ is constant. Indeed, substituting $k=$ const into $(2.33),(2.32)$, and (2.35), we obtain
\begin{aligned} &K(t)=k L(t), \quad I(t)=(\mu+\eta) K(t), \quad Q(t)=(\mu+\eta) K(t) / s \ &C(t)=Q(t)-I(t), \end{aligned}
i.e., all these functions increase with the same rate $\eta$ as the labor $L(t)=L_{0} \exp$ ( $\eta t)$. Therefore, to find steady states, we should assume $k(t)=$ const. Then $k^{\prime}(t)=0$ and the equation ( $2.36$ ) produces the equation
$$s f(k)=(\mu+\eta) k$$
for possible steady states $k \equiv$ const. Because $f(0)=0, f^{\prime}(k)>0, \lim {k \rightarrow 0} f^{\prime}(k)=\infty$, and $\lim {k \rightarrow \infty} f^{\prime}(k)=0$, the equation $(2.38$ ) has a unique solution $\hat{k}=\hat{k}(s)=$ const $>0$ for any given value $s>0$. The steady-state capital-labor ratio $\hat{k}(s)$ increases when the saving rate $s$ increases.

## 数学代写|数学生态学作业代写Mathematical Ecology代考|Static Optimization

For a given saving rate $s$, the steady-state consumption per capita $c=C / L$ is determined by the formula
$$c(s)=f(\hat{k}(s))-(\mu+h) \hat{k}(s)$$
where the corresponding steady-state capital-labor ratio $\hat{k}(s)$ is determined by (2.38). Because $f(0)=0, f^{\prime}(k)>0$ and $f^{\prime \prime}(k)<0$, the composite function (2.39) increases for smaller values of $s$ and decreases for larger $s$.

Then, we can determine the saving rate $s^{}=$ const and the corresponding steady-state $k^{}=\hat{k}\left(s^{}\right)$ that maximizes the consumption per capita (2.39): $$\max _{0} should satisfy$$
f^{\prime}\left(k^{*}\right)=\mu+\eta
$$The relation (2.40) is known as the golden rule of capital accumulation. It implies that the marginal product of capital should be equal to the sum of the depreciation and labor growth rates. After determining the optimal k^{} from (2.40), the corresponding golden-rule saving rate is found from (2.38) as$$ s^{}=(\mu+\eta) k^{} / f\left(k^{}\right)=k^{} f^{\prime}\left(k^{}\right) / f\left(k^{}\right), $$i.e., the optimal saving rate s^{} is equal to the output elasticity of the capital \varepsilon_{K} (2.6) for the corresponding k^{}. The formulas (2.40) and (2.41) for the optimal s^{} and k^{*} are known as the golden rule of economic growth. In the case of the Cobb-Douglas production function (2.22) F(K, L)= A K^{\alpha} L^{1-\alpha}, 0<\alpha<1, the function f(k)=A k^{\alpha} and the golden rule is$$
s^{}=\alpha, \quad k^{}=\left[A s^{} /(\mu+\eta)\right]^{1 /(1-\alpha)} . $$At the optimal steady state \left(s^{}, k^{}\right) and the given labor L(t)=\bar{L} \mathrm{e}^{m t}, the original variables Q(t), C(t), I(t), and K(t) of the model (2.31)-(2.35) grow with the given rate \eta as$$ \begin{aligned} K(t)=\bar{K}^{\eta t^{t}}, \quad I(t)=\bar{I} \mathrm{e}^{\eta t^{t}}, \quad Q(t)=\bar{Q} \mathrm{e}^{\eta t}, \quad C(t)=\bar{C} \mathrm{e}^{\eta t} \ \bar{K}=\bar{L} k^{}, \bar{I}=(\mu+\eta) \bar{L} k^{}, \quad \bar{Q}=\frac{(\mu+\eta) \bar{L} k^{}}{s} \
\bar{C}=\frac{(1-s)(\mu+\eta) \bar{L} k^{*}}{s}
\end{aligned}
$$The constants \bar{K}, \bar{I}, \bar{Q}, \bar{C} in exponential functions of the form (2.43) are often called in economics the level variables. In the case of constant labor L(t)=\bar{L} (i.e., \eta=0 ), the steady state is given by (2.44) and known as a stationary point. Because the aggregate output Q, consumption C, investment I and capital K increase with the same rate \eta as the exogenous labor L, the Solow-Swan model is classified in the economic theory as the exogenous growth model. ## 数学代写|数学生态学作业代写Mathematical Ecology代考|Optimization over Finite Horizon The Solow-Shell model is the Solow-Swan model (2.31)-(2.34) considered on a finite planning horizon [0, T] in the case when the saving rate s=I / Q depends on the time t and is endogenous [7]. To determine this rate, we consider the following one-sector optimization problem: • Maximize the present value$$
\int_{0}^{T} \mathrm{e}^{-r t} c(t) \mathrm{d} t
$$of the consumption per capita c=C / L over a given finite horizon [0, T], subject to (2.31)-(2.35) and certain initial and terminal conditions. In this problem, the given discount rate r>0 reflects the planner’s subjective rate of the decreasing utility of the output produced in more distant future. We still use the same aggregate variables of the Solow-Swan model: the output Q, consumption C, capital K, labor L, and investment I. For simplicity, let the labor L(t) be constant, that is, \eta=0 in (2.34). Switching the model (2.31)-(2.34) to the per capita variables k=K / L, q=Q / L, c=C / L, i=I / L, and excluding q, c, and i, the optimization problem under study becomes: • Find the function s(t), 0 \leq s(t) \leq 1, and the corresponding k(t), k(t) \geq 0, t \in[0, T], which maximize$$
\max {s, k} \int{0}^{T} \mathrm{e}^{-r t}(1-s(t)) f(k(t)) \mathrm{d} t
$$under the equality-constraint:$$
k^{\prime}(t)=s(t) f(k(t))-\mu k(t),
$$and the initial and terminal conditions:$$
$$The value of k(T) cannot be arbitrary because the economy will continue after the end of the planning period. The terminal condition k(T) \geq k_{\mathrm{T}} keeps a minimal acceptable level of capital at the end of the finite horizon. The problem (2.45)-(2.47) is an optimal control problem, in which the function s(t), t \in[0, T], is unknown (rather than the scalar s= const as in the static optimization of Sect. 2.2). In the optimal control terminology, the independent unknown function s(.) is referred to as the control variable and the corresponding dependent unknown k(.) is the state variable. ## 数学建模代写 ## 数学代写|数学生态学作业代写Mathematical Ecology代考|Steady-State Analysis 稳态分析的目标是找到可能的稳态，它可以是 • 静止轨迹（未知变量在时间上是恒定的）或 • 平衡的增长路径（所有变量以相同的恒定速率增长）。 稳态分析在经济学中起着重要作用，并且在数学上比完整的动态分析更简单。 让我们在模型中找到并分析可能的平衡增长路径(2.31)−(2.35). 很容易看出原来的变量问(吨),C(吨),一世(吨)， 和ķ(吨) 只有当资本-劳动力比率ķ(吨)是恒定的。确实，换ķ=常量成(2.33),(2.32), 和 (2.35), 我们得到 ķ(吨)=ķ大号(吨),一世(吨)=(μ+这)ķ(吨),问(吨)=(μ+这)ķ(吨)/s C(吨)=问(吨)−一世(吨), 即，所有这些功能都以相同的速度增加这作为劳动大号(吨)=大号0经验 ( 这吨). 因此，为了找到稳定状态，我们应该假设ķ(吨)=常量。然后ķ′(吨)=0和方程（2.36) 产生方程 sF(ķ)=(μ+这)ķ 对于可能的稳态ķ≡常量。因为F(0)=0,F′(ķ)>0,林ķ→0F′(ķ)=∞， 和林ķ→∞F′(ķ)=0, 方程(2.38) 有一个独特的解决方案ķ^=ķ^(s)=常量>0对于任何给定的值s>0. 稳态资本-劳动力比率ķ^(s)当储蓄率增加s增加。 ## 数学代写|数学生态学作业代写Mathematical Ecology代考|Static Optimization 对于给定的储蓄率s, 人均稳态消费C=C/大号由公式确定 C(s)=F(ķ^(s))−(μ+H)ķ^(s) 其中相应的稳态资本-劳动力比率ķ^(s)由 (2.38) 确定。因为F(0)=0,F′(ķ)>0和F′′(ķ)<0, 复合函数 (2.39) 对于较小的值会增加s并减少更大s. 然后，我们可以确定储蓄率s=const 和相应的稳态ķ=ķ^(s)最大化人均消费（2.39）：$$ \max _{0 }sH这在lds一种吨一世sF是F′(ķ∗)=μ+这\$

s=一种,ķ=[一种s/(μ+这)]1/(1−一种).处于最佳稳态(s,ķ)和给定的劳动大号(吨)=大号¯和米吨, 原始变量问(吨),C(吨),一世(吨)， 和ķ(吨)模型的 (2.31)-(2.35) 以给定的速率增长这作为ķ(吨)=ķ¯这吨吨,一世(吨)=一世¯和这吨吨,问(吨)=问¯和这吨,C(吨)=C¯和这吨 ķ¯=大号¯ķ,一世¯=(μ+这)大号¯ķ,问¯=(μ+这)大号¯ķs C¯=(1−s)(μ+这)大号¯ķ∗s

## 数学代写|数学生态学作业代写Mathematical Ecology代考|Optimization over Finite Horizon

Solow-Shell 模型是 Solow-Swan 模型(2.31)−(2.34)在有限的规划范围内考虑[0,吨]在储蓄率的情况下s=一世/问取决于时间吨并且是内生的[7]。为了确定这个比率，我们考虑以下单扇区优化问题：

• 最大化现值
∫0吨和−r吨C(吨)d吨
人均消费C=C/大号在给定的有限范围内[0,吨], 受限于 (2.31)-(2.35) 和某些初始和终止条件。

• 查找功能s(吨),0≤s(吨)≤1, 和对应的ķ(吨),ķ(吨)≥0, 吨∈[0,吨], 最大化
最大限度s,ķ∫0吨和−r吨(1−s(吨))F(ķ(吨))d吨
在等式约束下：
ķ′(吨)=s(吨)F(ķ(吨))−μķ(吨),
以及初始和终止条件：
ķ(0)=ķ0,ķ(吨)≥ķ吨
的价值ķ(吨)不能随意，因为经济将在计划期结束后继续。终端条件ķ(吨)≥ķ吨在有限期限结束时保持最低可接受的资本水平。

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