如果你也在 怎样代写凸优化Convex optimization 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。凸优化Convex optimization由于在大规模资源分配、信号处理和机器学习等领域的广泛应用，人们对凸优化的兴趣越来越浓厚。本书旨在解决凸优化问题的算法的最新和可访问的发展。

凸优化Convex optimization无约束可以很容易地用梯度下降(最陡下降的特殊情况)或牛顿方法解决，结合线搜索适当的步长;这些可以在数学上证明收敛速度很快，尤其是后一种方法。如果目标函数是二次函数，也可以使用KKT矩阵技术求解具有线性等式约束的凸优化(它推广到牛顿方法的一种变化，即使初始化点不满足约束也有效)，但通常也可以通过线性代数消除等式约束或解决对偶问题来解决。

statistics-lab™ 为您的留学生涯保驾护航 在代写凸优化Convex Optimization方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写凸优化Convex Optimization代写方面经验极为丰富，各种代写凸优化Convex Optimization相关的作业也就用不着说。

数学代写|凸优化作业代写Convex Optimization代考|Subgradient Methods with Iterate Averaging

If the stepsize $\alpha_k$ in the subgradient method
$$
x_{k+1}=P_X\left(x_k-\alpha_k g_k\right)
$$
is chosen to be large (such as constant or such that the condition $\sum_{k=0}^{\infty} \alpha_k^2<\infty$ is violated) the method may not converge. This exercise shows that by averaging the iterates of the method, we may obtain convergence with larger stepsizes. Let the optimal solution set $X^$ be nonempty, and assume that for some scalar $c$, we have $$ c \geq \sup \left{\left|g_k\right| \mid k=0,1, \ldots\right}, \quad \forall k \geq 0, $$ (cf. Assumption 3.2.1). Assume further that $\alpha_k$ is chosen according to $$ \alpha_k=\frac{\theta}{c \sqrt{k+1}}, \quad k=0,1, \ldots $$ where $\theta$ is a positive constant. Show that $$ f\left(\bar{x}k\right)-f^ \leq c\left(\frac{\min {x^* \in X^}\left|x_0-x^\right|^2}{2 \theta}+\theta \ln (k+2)\right) \frac{1}{\sqrt{k+1}}, \quad k=0,1, \ldots,
$$
where $\bar{x}k$ is the averaged iterate, generated according to $$ \bar{x}_k=\frac{\sum{\ell=0}^k \alpha_{\ell} x_{\ell}}{\sum_{\ell=0}^k \alpha_{\ell}}
$$

similar analysis applies to incremental and to stochastic subgradient methods.
Abbreviated proof: Denote
$$
\delta_k=\frac{1}{2} \min {x^* \in X^}\left|x_k-x^\right|^2 .
$$
Applying Prop. 3.2.2(a) with $y$ equal to the projection of $x_k$ onto $X^$, we obtain $$ \delta{k+1} \leq \delta_k-\alpha_k\left(f\left(x_k\right)-f^\right)+\frac{1}{2} \alpha_k^2 c^2
$$
Adding this inequality from 0 to $k$, and using the fact $\delta_{k+1} \geq 0$,
$$
\sum_{\ell=0}^k \alpha_{\ell}\left(f\left(x_k\right)-f^\right) \leq \delta_0+\frac{1}{2} c^2 \sum_{\ell=0}^k \alpha_{\ell}^2, $$ so by dividing with $\sum_{\ell=0}^k \alpha_{\ell}$, $$ \frac{\sum_{\ell=0}^k \alpha_{\ell} f\left(x_k\right)}{\sum_{\ell=0}^k \alpha_{\ell}}-f^ \leq \frac{\delta_0+\frac{1}{2} c^2 \sum_{\ell=0}^k \alpha_{\ell}^2}{\sum_{\ell=0}^k \alpha_{\ell}}
$$

数学代写|凸优化作业代写Convex Optimization代考|Modified Dynamic Stepsize Rules

Consider the subgradient method
$$
x_{k+1}=P_X\left(x_k-\alpha_k g_k\right)
$$
with the stepsize chosen according to one of the two rules
$$
\alpha_k=\frac{f\left(x_k\right)-f_k}{\max \left{\gamma,\left|g_k\right|^2\right}} \quad \text { or } \quad \alpha_k=\min \left{\gamma, \frac{f\left(x_k\right)-f_k}{\left|g_k\right|^2}\right}
$$

where $\gamma$ is a fixed positive scalar and $f_k$ is given by the dynamic adjustment procedure (3.22)-(3.23). Show that the convergence result of Prop. 3.2.8 still holds. Abbreviated Proof: We proceed by contradiction, as in the proof of Prop. 3.2.8. From Prop. 3.2.2(a) with $y=\bar{y}$, we have for all $k \geq \bar{k}$,
$$
\begin{aligned}
\left|x_{k+1}-\bar{y}\right|^2 & \leq\left|x_k-\bar{y}\right|^2-2 \alpha_k\left(f\left(x_k\right)-f(\bar{y})\right)+\alpha_k^2\left|g_k\right|^2 \
& \leq\left|x_k-\bar{y}\right|^2-2 \alpha_k\left(f\left(x_k\right)-f(\bar{y})\right)+\alpha_k\left(f\left(x_k\right)-f_k\right) \
& =\left|x_k-\bar{y}\right|^2-\alpha_k\left(f\left(x_k\right)-f_k\right)-2 \alpha_k\left(f_k-f(\bar{y})\right) \
& \leq\left|x_k-\bar{y}\right|^2-\alpha_k\left(f\left(x_k\right)-f_k\right) .
\end{aligned}
$$
Hence $\left{x_k\right}$ is bounded, which implies that $\left{g_k\right}$ is also bounded (cf. Prop. 3.1.2). Let $\bar{c}$ be such that $\left|g_k\right| \leq \bar{c}$ for all $k$. Assume that $\alpha_k$ is chosen according to the first rule in Eq. (3.47). Then from the preceding relation we have for all $k \geq \bar{k}$,
$$
\left|x_{k+1}-\bar{y}\right|^2 \leq\left|x_k-\bar{y}\right|^2-\frac{\delta^2}{\max \left{\gamma, \bar{c}^2\right}} .
$$
As in the proof of Prop. 3.2.8, this leads to a contradiction and the result follows. The proof is similar if $\alpha_k$ is chosen according to the second rule in Eq. (3.47).

凸优化代写

数学代写|凸优化作业代写Convex Optimization代考|Subgradient Methods with Iterate Averaging

如果步长为$\alpha_k$在子梯度法中
$$
x_{k+1}=P_X\left(x_k-\alpha_k g_k\right)
$$
选择较大(如常数或违反$\sum_{k=0}^{\infty} \alpha_k^2<\infty$条件)的方法可能不收敛。这个练习表明，通过平均方法的迭代，我们可以在较大的步长下获得收敛性。令最优解集$X^$非空，并假设对于某个标量$c$，我们有$$ c \geq \sup \left{\left|g_k\right| \mid k=0,1, \ldots\right}, \quad \forall k \geq 0, $$(参见假设3.2.1)。进一步假设$\alpha_k$是根据$$ \alpha_k=\frac{\theta}{c \sqrt{k+1}}, \quad k=0,1, \ldots $$选择的，其中$\theta$是一个正常数。展示一下$$ f\left(\bar{x}k\right)-f^ \leq c\left(\frac{\min {x^* \in X^}\left|x_0-x^\right|^2}{2 \theta}+\theta \ln (k+2)\right) \frac{1}{\sqrt{k+1}}, \quad k=0,1, \ldots,
$$
其中$\bar{x}k$是根据生成的平均迭代 $$ \bar{x}k=\frac{\sum{\ell=0}^k \alpha{\ell} x_{\ell}}{\sum_{\ell=0}^k \alpha_{\ell}}
$$

类似的分析也适用于增量法和随机次梯度法。
缩写证明:
$$
\delta_k=\frac{1}{2} \min {x^* \in X^}\left|x_k-x^\right|^2 .
$$
应用Prop. 3.2.2(a)，令$y$等于$x_k$在$X^$上的投影，我们得到$$ \delta{k+1} \leq \delta_k-\alpha_k\left(f\left(x_k\right)-f^\right)+\frac{1}{2} \alpha_k^2 c^2
$$
把不等式从0加到$k$，利用$\delta_{k+1} \geq 0$这个事实，
$$
\sum_{\ell=0}^k \alpha_{\ell}\left(f\left(x_k\right)-f^\right) \leq \delta_0+\frac{1}{2} c^2 \sum_{\ell=0}^k \alpha_{\ell}^2, $$除以$\sum_{\ell=0}^k \alpha_{\ell}$， $$ \frac{\sum_{\ell=0}^k \alpha_{\ell} f\left(x_k\right)}{\sum_{\ell=0}^k \alpha_{\ell}}-f^ \leq \frac{\delta_0+\frac{1}{2} c^2 \sum_{\ell=0}^k \alpha_{\ell}^2}{\sum_{\ell=0}^k \alpha_{\ell}}
$$

数学代写|凸优化作业代写Convex Optimization代考|Modified Dynamic Stepsize Rules

考虑次梯度法
$$
x_{k+1}=P_X\left(x_k-\alpha_k g_k\right)
$$
步长根据两个规则之一选择
$$
\alpha_k=\frac{f\left(x_k\right)-f_k}{\max \left{\gamma,\left|g_k\right|^2\right}} \quad \text { or } \quad \alpha_k=\min \left{\gamma, \frac{f\left(x_k\right)-f_k}{\left|g_k\right|^2}\right}
$$

其中$\gamma$为固定正标量，$f_k$由式(3.22)-(3.23)的动态调整过程给出。证明Prop. 3.2.8的收敛性结果仍然成立。简证:我们以反证法进行，如提案3.2.8的证明。根据提案3.2.2(a)中$y=\bar{y}$，我们得到所有$k \geq \bar{k}$，
$$
\begin{aligned}
\left|x_{k+1}-\bar{y}\right|^2 & \leq\left|x_k-\bar{y}\right|^2-2 \alpha_k\left(f\left(x_k\right)-f(\bar{y})\right)+\alpha_k^2\left|g_k\right|^2 \
& \leq\left|x_k-\bar{y}\right|^2-2 \alpha_k\left(f\left(x_k\right)-f(\bar{y})\right)+\alpha_k\left(f\left(x_k\right)-f_k\right) \
& =\left|x_k-\bar{y}\right|^2-\alpha_k\left(f\left(x_k\right)-f_k\right)-2 \alpha_k\left(f_k-f(\bar{y})\right) \
& \leq\left|x_k-\bar{y}\right|^2-\alpha_k\left(f\left(x_k\right)-f_k\right) .
\end{aligned}
$$
因此$\left{x_k\right}$是有界的，这就意味着$\left{g_k\right}$也是有界的(参见Prop 3.1.2)。让$\bar{c}$为所有$k$成为$\left|g_k\right| \leq \bar{c}$。假设根据式(3.47)中的第一条规则选择$\alpha_k$。从前面的关系式中我们得到所有$k \geq \bar{k}$，
$$
\left|x_{k+1}-\bar{y}\right|^2 \leq\left|x_k-\bar{y}\right|^2-\frac{\delta^2}{\max \left{\gamma, \bar{c}^2\right}} .
$$
正如Prop. 3.2.8的证明一样，这就引出了一个矛盾，结果就出来了。如果根据式(3.47)中的第二条规则选择$\alpha_k$，则证明是类似的。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

如果你也在 怎样代写凸优化Convex optimization 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。凸优化Convex optimization由于在大规模资源分配、信号处理和机器学习等领域的广泛应用，人们对凸优化的兴趣越来越浓厚。本书旨在解决凸优化问题的算法的最新和可访问的发展。

凸优化Convex optimization无约束可以很容易地用梯度下降(最陡下降的特殊情况)或牛顿方法解决，结合线搜索适当的步长;这些可以在数学上证明收敛速度很快，尤其是后一种方法。如果目标函数是二次函数，也可以使用KKT矩阵技术求解具有线性等式约束的凸优化(它推广到牛顿方法的一种变化，即使初始化点不满足约束也有效)，但通常也可以通过线性代数消除等式约束或解决对偶问题来解决。

statistics-lab™ 为您的留学生涯保驾护航 在代写凸优化Convex Optimization方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写凸优化Convex Optimization代写方面经验极为丰富，各种代写凸优化Convex Optimization相关的作业也就用不着说。

数学代写|凸优化作业代写Convex Optimization代考|Connection with Incremental Subgradient Methods

We discussed in Section 2.1.5 incremental variants of gradient methods, which apply to minimization over a closed convex set $X$ of an additive cost function of the form
$$
f(x)=\sum_{i=1}^m f_i(x),
$$
where the functions $f_i: \Re^n \mapsto \Re$ are differentiable. Incremental variants of the subgradient method are also possible in the case where the $f_i$ are nondifferentiable but convex. The idea is to sequentially take steps along the subgradients of the component functions $f_i$, with intermediate adjustment of $x$ after processing each $f_i$. We simply use an arbitrary subgradient of $f_i$ at a point where $f_i$ is nondifferentiable, in place of the gradient that would be used if $f_i$ were differentiable at that point.

Incremental methods are particularly interesting when the number of cost terms $m$ is very large. Then a full subgradient step is very costly, and one hopes to make progress with approximate but much cheaper incremental steps. We will discuss in detail incremental subgradient methods and their combinations with other methods, such as incremental proximal methods, in Section 6.4. In this section we will discuss the most common type of incremental subgradient method, and highlight its connection with the $\epsilon$-subgradient method.

Let us consider the minimization of $\sum_{i=1}^m f_i$ over $x \in X$, for the case where each $f_i$ is a convex real-valued function. Similar to the incremental gradient methods of Section 2.1.5, we view an iteration as a cycle of $m$ subiterations. If $x_k$ is the vector obtained after $k$ cycles, the vector $x_{k+1}$ obtained after one more cycle is
$$
x_{k+1}=\psi_{m, k},
$$
where starting with $\psi_{0, k}=x_k$, we obtain $\psi_{m, k}$ after the $m$ steps
$$
\psi_{i, k}=P_X\left(\psi_{i-1, k}-\alpha_k g_{i, k}\right), \quad i=1, \ldots, m,
$$
with $g_{i, k}$ being an arbitrary subgradient of $f_i$ at $\psi_{i-1, k}$.

数学代写|凸优化作业代写Convex Optimization代考|NOTES, SOURCES, AND EXERCISES

Section 3.1: Subgradients are central in the work of Fenchel [Fen51]. The original theorem by Danskin [Dan67] provides a formula for the directional derivative of the maximum of a (not necessarily convex) directionally differentiable function. When adapted to a convex function $f$, this formula yields Eq. (3.10) for the subdifferential of $f$; see Exercise 3.5.

Another important subdifferential formula relates to the subgradients of an expected value function
$$
f(x)=E{F(x, \omega)},
$$

where $\omega$ is a random variable taking values in a set $\Omega$, and $F(\cdot, \omega): \Re^n \mapsto \Re$ is a real-valued convex function such that $f$ is real-valued (note that $f$ is easily verified to be convex). If $\omega$ takes a finite number of values with probabilities $p(\omega)$, then the formulas
$$
f^{\prime}(x ; d)=E\left{F^{\prime}(x, \omega ; d)\right}, \quad \partial f(x)=E{\partial F(x, \omega)},
$$
hold because they can be written in terms of finite sums as
$$
f^{\prime}(x ; d)=\sum_{\omega \in \Omega} p(\omega) F^{\prime}(x, \omega ; d), \quad \partial f(x)=\sum_{\omega \in \Omega} p(\omega) \partial F(x, \omega),
$$
so Prop. 3.1.3(b) applies. However, the formulas (3.32) hold even in the case where $\Omega$ is uncountably infinite, with appropriate mathematical interpretation of the integral of set-valued functions $E{\partial F(x, \omega)}$ as the set of integrals
$$
\int_{\omega \in \Omega} g(x, \omega) d P(\omega)
$$
where $g(x, \omega) \in \partial F(x, \omega), \omega \in \Omega$ (measurability issues must be addressed in this context). For a formal proof and analysis, see the author’s papers [Ber72], [Ber73], which also provide a necessary and sufficient condition for $f$ to be differentiable, even when $F(\cdot, \omega)$ is not. In this connection, it is important to note that the integration over $\omega$ in Eq. (3.33) may smooth out the nondifferentiabilities of $F(\cdot, \omega)$ if $\omega$ is a “continuous” random variable. This property can be used in turn in algorithms, including schemes that bring to bear the methodology of differentiable optimization; see e.g., Yousefian, Nedić, and Shanbhag [YNS10], [YNS12], Agarwal and Duchi [AgD11], Duchi, Bartlett, and Wainwright [DBW12], Brown and Smith [BrS13], Abernethy et al. [ALS14], and Jiang and Zhang [JiZ14].

凸优化代写

数学代写|凸优化作业代写Convex Optimization代考|Connection with Incremental Subgradient Methods

我们在2.1.5节中讨论了梯度方法的增量变体，它适用于在闭凸集$X$上最小化形式为的附加代价函数
$$
f(x)=\sum_{i=1}^m f_i(x),
$$
这里的函数$f_i: \Re^n \mapsto \Re$是可微的。在$f_i$不可微但是凸的情况下，子梯度方法的增量变体也是可能的。其思想是沿着组件函数$f_i$的子梯度依次采取步骤，在处理每个$f_i$之后对$x$进行中间调整。我们只需在$f_i$不可微的一点上使用$f_i$的任意子梯度，来代替$f_i$在该点可微时使用的梯度。

当成本项$m$的数量非常大时，增量方法特别有趣。然后，一个完整的次梯度步骤是非常昂贵的，人们希望通过近似但更便宜的增量步骤取得进展。我们将在第6.4节详细讨论增量亚梯度方法及其与其他方法的组合，例如增量近端方法。在本节中，我们将讨论最常见的增量子梯度方法类型，并强调其与$\epsilon$ -subgradient方法的联系。

让我们考虑$\sum_{i=1}^m f_i$ / $x \in X$的最小化，对于每个$f_i$都是凸实值函数的情况。与第2.1.5节的增量梯度方法类似，我们将迭代视为$m$子迭代的循环。若$x_k$为$k$次循环后得到的矢量，则再经过一次循环后得到的矢量$x_{k+1}$为
$$
x_{k+1}=\psi_{m, k},
$$
从$\psi_{0, k}=x_k$开始，我们在$m$步骤之后得到$\psi_{m, k}$
$$
\psi_{i, k}=P_X\left(\psi_{i-1, k}-\alpha_k g_{i, k}\right), \quad i=1, \ldots, m,
$$
其中$g_{i, k}$是$f_i$ at $\psi_{i-1, k}$的任意子梯度。

数学代写|凸优化作业代写Convex Optimization代考|NOTES, SOURCES, AND EXERCISES

第3.1节:次梯度是Fenchel [Fen51]工作的核心。Danskin [Dan67]的原始定理提供了一个方向可微函数(不一定是凸的)最大值的方向导数公式。当适用于凸函数$f$时，该公式为$f$的次微分产生Eq. (3.10);参见练习3.5。

另一个重要的次微分公式与期望值函数的次梯度有关
$$
f(x)=E{F(x, \omega)},
$$

其中$\omega$是在集合$\Omega$中取值的随机变量，$F(\cdot, \omega): \Re^n \mapsto \Re$是实值凸函数，因此$f$是实值(注意$f$很容易被验证为凸)。如果$\omega$取有限个数的值，概率为$p(\omega)$，则公式
$$
f^{\prime}(x ; d)=E\left{F^{\prime}(x, \omega ; d)\right}, \quad \partial f(x)=E{\partial F(x, \omega)},
$$
因为它们可以写成有限和的形式
$$
f^{\prime}(x ; d)=\sum_{\omega \in \Omega} p(\omega) F^{\prime}(x, \omega ; d), \quad \partial f(x)=\sum_{\omega \in \Omega} p(\omega) \partial F(x, \omega),
$$
因此，第3.1.3(b)号提案适用。然而，公式(3.32)即使在$\Omega$是不可数无穷的情况下也成立，将集值函数的积分$E{\partial F(x, \omega)}$适当地解释为积分集
$$
\int_{\omega \in \Omega} g(x, \omega) d P(\omega)
$$
其中$g(x, \omega) \in \partial F(x, \omega), \omega \in \Omega$(可度量性问题必须在此上下文中解决)。对于正式的证明和分析，参见作者的论文[Ber72]， [Ber73]，它们也提供了$f$可微的充分必要条件，即使$F(\cdot, \omega)$不可微。在这方面，重要的是要注意，如果$\omega$是一个“连续”随机变量，则Eq.(3.33)中对$\omega$的积分可以平滑$F(\cdot, \omega)$的不可微性。这个性质可以依次用于算法中，包括采用可微优化方法的方案;参见Yousefian, nedidic, and Shanbhag [YNS10]， [YNS12]， Agarwal and Duchi [AgD11]， Duchi, Bartlett, and Wainwright [DBW12]， Brown and Smith [BrS13]， Abernethy et al. [ALS14]， Jiang and Zhang [JiZ14]。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

如果你也在 怎样代写凸优化Convex optimization 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。凸优化Convex optimization由于在大规模资源分配、信号处理和机器学习等领域的广泛应用，人们对凸优化的兴趣越来越浓厚。本书旨在解决凸优化问题的算法的最新和可访问的发展。

凸优化Convex optimization无约束可以很容易地用梯度下降(最陡下降的特殊情况)或牛顿方法解决，结合线搜索适当的步长;这些可以在数学上证明收敛速度很快，尤其是后一种方法。如果目标函数是二次函数，也可以使用KKT矩阵技术求解具有线性等式约束的凸优化(它推广到牛顿方法的一种变化，即使初始化点不满足约束也有效)，但通常也可以通过线性代数消除等式约束或解决对偶问题来解决。

statistics-lab™ 为您的留学生涯保驾护航 在代写凸优化Convex Optimization方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写凸优化Convex Optimization代写方面经验极为丰富，各种代写凸优化Convex Optimization相关的作业也就用不着说。

数学代写|凸优化作业代写Convex Optimization代考|Linear Convergence Rate of Incremental Gradient Method

This exercise quantifies the rate of convergence of the incremental gradient method to the “region of confusion” (cf. Fig. 2.1.11), for any order of processing the additive cost components, assuming these components are positive definite quadratic. Consider the incremental gradient method
$$
x_{k+1}=x_k-\alpha \nabla f_k\left(x_k\right) \quad k=0,1, \ldots,
$$
where $f_0, f_1, \ldots$, are quadratic functions with eigenvalues lying within some interval $[\gamma, \Gamma]$, where $\gamma>0$. Suppose that for a given $\epsilon>0$, there is a vector $x^$ such that $$ \left|\nabla f_k\left(x^\right)\right| \leq \epsilon, \quad \forall k=0,1, \ldots
$$

Show that for all $\alpha$ with $0<\alpha \leq 2 /(\gamma+\Gamma)$, the generated sequence $\left{x_k\right}$ converges to a $2 \epsilon / \gamma$-neighborhood of $x^$, i.e., $$ \limsup {k \rightarrow \infty}\left|x_k-x^\right| \leq \frac{2 \epsilon}{\gamma} . $$ Moreover the rate of convergence to this neighborhood is linear, in the sense that $$ \left|x_k-x^\right|>\frac{2 \epsilon}{\gamma} \quad \Rightarrow \quad\left|x{k+1}-x^\right|<\left(1-\frac{\alpha \gamma}{2}\right)\left|x_k-x^\right|, $$ while $$ \left|x_k-x^\right| \leq \frac{2 \epsilon}{\gamma} \quad \Rightarrow \quad\left|x_{k+1}-x^\right| \leq \frac{2 \epsilon}{\gamma} . $$ Hint: Let $f_k(x)=\frac{1}{2} x^{\prime} Q_k x-b_k^{\prime} x$, where $Q_k$ is positive definite symmetric, and write $$ x_{k+1}-x^=\left(I-\alpha Q_k\right)\left(x_k-x^\right)-\alpha \nabla f_k\left(x^\right) .
$$
For other related convergence rate results, see [NeB00] and [Sch14a].

数学代写|凸优化作业代写Convex Optimization代考|Proximal Gradient Method, £1-Regularization, and the Shrinkage Operation

The proximal gradient iteration (2.27) is well suited for problems involving a nondifferentiable function component that is convenient for a proximal iteration. This exercise considers the important case of the $\ell_1$ norm. Consider the problem
$$
\begin{aligned}
& \operatorname{minimize} f(x)+\gamma|x|_1 \
& \text { subject to } x \in \Re^n,
\end{aligned}
$$
where $f: \Re^n \mapsto \Re$ is a differentiable convex function, $|\cdot|_1$ is the $\ell_1$ norm, and $\gamma>0$. The proximal gradient iteration is given by the gradient step
$$
z_k=x_k-\alpha \nabla f\left(x_k\right)
$$
followed by the proximal step
$$
x_{k+1} \in \arg \min {x \in \Re^n}\left{\gamma|x|_1+\frac{1}{2 \alpha}\left|x-z_k\right|^2\right} $$ [cf. Eq. (2.28)]. Show that the proximal step can be performed separately for each coordinate $x^i$ of $x$, and is given by the so-called shrinkage operation: $$ x{k+1}^i=\left{\begin{array}{ll}
z_k^i-\alpha \gamma & \text { if } z_k^i>\alpha \gamma, \
0 & \text { if }\left|z_k^i\right| \leq \alpha \gamma, \
z_k^i+\alpha \gamma & \text { if } z_k^i<-\alpha \gamma,
\end{array} \quad i=1, \ldots, n .\right.
$$
Note: Since the shrinkage operation tends to set many coordinates $x_{k+1}^i$ to 0 , it tends to produce “sparse” iterates.

凸优化代写

数学代写|凸优化作业代写Convex Optimization代考|Linear Convergence Rate of Incremental Gradient Method

这个练习量化了增量梯度法在“混乱区域”的收敛速度(参见图2.1.11)，对于任何处理附加成本分量的顺序，假设这些分量是正定的二次元。考虑增量梯度法
$$
x_{k+1}=x_k-\alpha \nabla f_k\left(x_k\right) \quad k=0,1, \ldots,
$$
式中$f_0, f_1, \ldots$为特征值位于某区间$[\gamma, \Gamma]$内的二次函数，式中$\gamma>0$。假设对于给定的$\epsilon>0$，有一个向量$x^$满足 $$ \left|\nabla f_k\left(x^\right)\right| \leq \epsilon, \quad \forall k=0,1, \ldots
$$

证明对于所有$\alpha$和$0<\alpha \leq 2 /(\gamma+\Gamma)$，生成的序列$\left{x_k\right}$收敛到$x^$的$2 \epsilon / \gamma$邻域，即$$ \limsup {k \rightarrow \infty}\left|x_k-x^\right| \leq \frac{2 \epsilon}{\gamma} . $$，并且收敛到该邻域的速度是线性的，即$$ \left|x_k-x^\right|>\frac{2 \epsilon}{\gamma} \quad \Rightarrow \quad\left|x{k+1}-x^\right|<\left(1-\frac{\alpha \gamma}{2}\right)\left|x_k-x^\right|, $$和$$ \left|x_k-x^\right| \leq \frac{2 \epsilon}{\gamma} \quad \Rightarrow \quad\left|x_{k+1}-x^\right| \leq \frac{2 \epsilon}{\gamma} . $$提示:设$f_k(x)=\frac{1}{2} x^{\prime} Q_k x-b_k^{\prime} x$，其中$Q_k$是正定对称的，并写$$ x_{k+1}-x^=\left(I-\alpha Q_k\right)\left(x_k-x^\right)-\alpha \nabla f_k\left(x^\right) .
$$
其他相关的收敛速率结果参见[NeB00]和[Sch14a]。

数学代写|凸优化作业代写Convex Optimization代考|Proximal Gradient Method, £1-Regularization, and the Shrinkage Operation

近端梯度迭代(2.27)非常适合于涉及不可微函数分量的问题，这便于近端迭代。这个练习考虑了$\ell_1$规范的重要情况。考虑这个问题
$$
\begin{aligned}
& \operatorname{minimize} f(x)+\gamma|x|1 \ & \text { subject to } x \in \Re^n, \end{aligned} $$ 其中$f: \Re^n \mapsto \Re$为可微凸函数，$|\cdot|_1$为$\ell_1$范数，$\gamma>0$。近端梯度迭代由梯度步长给出 $$ z_k=x_k-\alpha \nabla f\left(x_k\right) $$ 接着是近端步骤 $$ x{k+1} \in \arg \min {x \in \Re^n}\left{\gamma|x|1+\frac{1}{2 \alpha}\left|x-z_k\right|^2\right} $$[参见式(2.28)]。表明可以对$x$的每个坐标$x^i$分别执行近端步骤，并由所谓的收缩操作:$$ x{k+1}^i=\left{\begin{array}{ll} z_k^i-\alpha \gamma & \text { if } z_k^i>\alpha \gamma, \ 0 & \text { if }\left|z_k^i\right| \leq \alpha \gamma, \ z_k^i+\alpha \gamma & \text { if } z_k^i<-\alpha \gamma, \end{array} \quad i=1, \ldots, n .\right. $$给出 注意:由于收缩操作倾向于将许多坐标$x{k+1}^i$设置为0，因此它倾向于产生“稀疏”迭代。

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如果你也在 怎样代写凸优化Convex optimization 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。凸优化Convex optimization凸优化是数学优化的一个子领域，研究的是凸集上凸函数最小化的问题。许多类凸优化问题允许采用多项式时间算法，而数学优化一般来说是NP-hard。

凸优化Convex optimization是数学优化的一个子领域，研究的是凸集上凸函数最小化的问题。许多类别的凸优化问题允许采用多项式时间算法，而数学优化一般来说是NP困难的。凸优化在许多学科中都有应用，如自动控制系统、估计和信号处理、通信和网络、电子电路设计、数据分析和建模、金融、统计（最佳实验设计）、和结构优化，其中近似概念被证明是有效的。

statistics-lab™ 为您的留学生涯保驾护航 在代写凸优化Convex Optimization方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写凸优化Convex Optimization代写方面经验极为丰富，各种代写凸优化Convex Optimization相关的作业也就用不着说。

数学代写|凸优化作业代写Convex Optimization代考|Armijo/Backtracking Stepsize Rule

Consider minimization of a continuously differentiable function $f: \Re^n \mapsto \Re$, using the iteration
$$
x_{k+1}=x_k+\alpha_k d_k
$$
where $d_k$ is a descent direction. Given fixed scalars $\beta$, and $\sigma$, with $0<\beta<1$, $0<\sigma<1$, and $s_k$ with $\inf _{k \geq 0} s_k>0$, the stepsize $\alpha_k$ is determined as follows: we set $\alpha_k=\beta^{m_k} s_k$, where $m_k$ is the first nonnegative integer $m$ for which
$$
f\left(x_k\right)-f\left(x_k+\beta^m s_k d_k\right) \geq-\sigma \beta^m s_k \nabla f\left(x_k\right)^{\prime} d_k
$$
Assume that there exist positive scalars $c_1, c_2$ such that for all $k$ we have
$$
c_1\left|\nabla f\left(x_k\right)\right|^2 \leq-\nabla f\left(x_k\right)^{\prime} d_k, \quad\left|d_k\right|^2 \leq c_2\left|\nabla f\left(x_k\right)\right|^2
$$
(a) Show that the stepsize $\alpha_k$ is well-defined, i.e., that it will be determined after a finite number of reductions if $\nabla f\left(x_k\right) \neq 0$. Proof: We have for all $s>0$
$$
f\left(x_k+s d_k\right)-f\left(x_k\right)=s \nabla f\left(x_k\right)^{\prime} d_k+o(s)
$$
Thus the test for acceptance of a stepsize $s>0$ is written as
$$
s \nabla f\left(x_k\right)^{\prime} d_k+o(s) \leq \sigma s \nabla f\left(x_k\right)^{\prime} d_k,
$$
or using Eq. (2.72),
$$
\frac{o(s)}{s} \leq(1-\sigma) c_1\left|\nabla f\left(x_k\right)\right|^2,
$$
which is satisfied for $s$ in some interval $\left(0, \bar{s}_k\right]$. Thus the test will be passed for all $m$ for which $\beta^m s_k \leq \bar{s}_k$.

Show that every limit point $\bar{x}$ of the generated sequence $\left{x_k\right}$ satisfies $\nabla f(\bar{x})=0$. Proof: Assume, to arrive at a contradiction, that there is a subsequence $\left{x_k\right}_{\mathcal{K}}$ that converges to some $\bar{x}$ with $\nabla f(\bar{x}) \neq 0$. Since $\left{f\left(x_k\right)\right}$ is monotonically nonincreasing, $\left{f\left(x_k\right)\right}$ either converges to a finite value or diverges to $-\infty$. Since $f$ is continuous, $f(\bar{x})$ is a limit point of $\left{f\left(x_k\right)\right}$, so it follows that the entire sequence $\left{f\left(x_k\right)\right}$ converges to $f(\bar{x})$. Hence,
$$
f\left(x_k\right)-f\left(x_{k+1}\right) \rightarrow 0
$$
By the definition of the Armijo rule and the descent property $\nabla f\left(x_k\right)^{\prime} d_k \leq$ 0 of the direction $d_k$, we have
$$
f\left(x_k\right)-f\left(x_{k+1}\right) \geq-\sigma \alpha_k \nabla f\left(x_k\right)^{\prime} d_k \geq 0,
$$
so by combining the preceding two relations,
$$
\alpha_k \nabla f\left(x_k\right)^{\prime} d_k \rightarrow 0
$$
From the left side of Eq. (2.72) and the hypothesis $\nabla f(\bar{x}) \neq 0$, it follows that
$$
\limsup {\substack{k \rightarrow \infty \ k \in \mathcal{K}}} \nabla f\left(x_k\right)^{\prime} d_k<0 $$ which together with Eq. (2.73) implies that $$ \left{\alpha_k\right}{\mathcal{K}} \rightarrow 0 .
$$
which together with Eq. (2.73) implies that
$$
\left{\alpha_k\right}_{\mathcal{K}} \rightarrow 0
$$
Since $s_k$, the initial trial value for $\alpha_k$, is bounded away from $0, s_k$ will be reduced at least once for all $k \in \mathcal{K}$ that are greater than some iteration index $\bar{k}$. Thus we must have for all $k \in \mathcal{K}$ with $k>\bar{k}$
$$
f\left(x_k\right)-f\left(x_k+\left(\alpha_k / \beta\right) d_k\right)<-\sigma\left(\alpha_k / \beta\right) \nabla f\left(x_k\right)^{\prime} d_k
$$

数学代写|凸优化作业代写Convex Optimization代考|Convergence of Steepest Descent to a Single Limit

Let $f: \Re^n \mapsto \Re$ be a differentiable convex function, and assume that for some $L>0$, we have
$$
|\nabla f(x)-\nabla f(y)| \leq L|x-y|, \quad \forall x, y \in \Re^n .
$$
Let $X^$ be the set of minima of $f$, and assume that $X^$ is nonempty. Consider the steepest descent method
$$
x_{k+1}=x_k-\alpha_k \nabla f\left(x_k\right)
$$
Show that $\left{x_k\right}$ converges to a minimizing point of $f$ under each of the following two stepsize rule conditions:
(i) For some $\epsilon>0$, we have
$$
\epsilon \leq \alpha_k \leq \frac{2(1-\epsilon)}{L}, \quad \forall k .
$$
(ii) $\alpha_k \rightarrow 0$ and $\sum_{k=0}^{\infty} \alpha_k=\infty$.
Notes: The original source [BGI95] also shows convergence to a single limit for a variant of the Armijo rule. This should be contrasted with a result of [Gon00], which shows that the steepest descent method with the exact line minimization rule may produce a sequence with multiple limit points (all of which are of course optimal), even for a convex cost function. There is also a “local capture” theorem that applies to gradient methods for nonconvex continuously differentiable cost functions $f$ and an isolated local minimum of $f$ (a local minimum $x^$ that is unique within a neighborhood of $x^$ ). Under mild conditions it asserts that there is an open sphere $S_{x^}$ centered at $x^$ such that once the generated sequence $\left{x_k\right}$ enters $S_{x^}$, it converges to $x^$ (see [Ber82a], Prop. 1.12, or [Ber99], Prop. 1.2.5 and the references given there). Abbreviated Proof: Consider the stepsize rule (i). From the descent inequality (Exercise 2.2), we have for all $k$
$$
f\left(x_{k+1}\right) \leq f\left(x_k\right)-\alpha_k\left(1-\frac{\alpha_k L}{2}\right)\left|\nabla f\left(x_k\right)\right|^2 \leq f\left(x_k\right)-\epsilon^2\left|\nabla f\left(x_k\right)\right|^2
$$
so $\left{f\left(x_k\right)\right}$ is monotonically nonincreasing and converges. Adding the preceding relation for all values of $k$ and taking the limit as $k \rightarrow \infty$, we obtain for all $x^* \in X^$, $$ f\left(x^\right) \leq f\left(x_0\right)-\epsilon^2 \sum_{k=0}^{\infty}\left|\nabla f\left(x_k\right)\right|^2
$$

凸优化代写

数学代写|凸优化作业代写Convex Optimization代考|Armijo/Backtracking Stepsize Rule

考虑最小化一个连续可微函数$f: \Re^n \mapsto \Re$，使用迭代
$$
x_{k+1}=x_k+\alpha_k d_k
$$
其中$d_k$为下降方向。给定固定标量$\beta$, $\sigma$, $0<\beta<1$, $0<\sigma<1$, $s_k$, $\inf _{k \geq 0} s_k>0$，步长$\alpha_k$确定如下:我们设置$\alpha_k=\beta^{m_k} s_k$，其中$m_k$是第一个非负整数$m$
$$
f\left(x_k\right)-f\left(x_k+\beta^m s_k d_k\right) \geq-\sigma \beta^m s_k \nabla f\left(x_k\right)^{\prime} d_k
$$
假设存在正标量$c_1, c_2$对于所有的$k$
$$
c_1\left|\nabla f\left(x_k\right)\right|^2 \leq-\nabla f\left(x_k\right)^{\prime} d_k, \quad\left|d_k\right|^2 \leq c_2\left|\nabla f\left(x_k\right)\right|^2
$$
(a)证明步长$\alpha_k$是定义良好的，即，如果$\nabla f\left(x_k\right) \neq 0$，它将在有限次缩减后确定。证明:我们有所有$s>0$
$$
f\left(x_k+s d_k\right)-f\left(x_k\right)=s \nabla f\left(x_k\right)^{\prime} d_k+o(s)
$$
因此，步长$s>0$的可接受性测试写为
$$
s \nabla f\left(x_k\right)^{\prime} d_k+o(s) \leq \sigma s \nabla f\left(x_k\right)^{\prime} d_k,
$$
或用式(2.72)，
$$
\frac{o(s)}{s} \leq(1-\sigma) c_1\left|\nabla f\left(x_k\right)\right|^2,
$$
它在$\left(0, \bar{s}_k\right]$区间内满足$s$。因此，测试将通过所有$m$，其中$\beta^m s_k \leq \bar{s}_k$。

证明生成序列$\left{x_k\right}$的每个极限点$\bar{x}$满足$\nabla f(\bar{x})=0$。证明:为了得到一个矛盾，假设有一个子序列$\left{x_k\right}{\mathcal{K}}$与$\nabla f(\bar{x}) \neq 0$收敛到某个$\bar{x}$。由于$\left{f\left(x_k\right)\right}$是单调非递增的，因此$\left{f\left(x_k\right)\right}$收敛于有限值或发散到$-\infty$。因为$f$是连续的，$f(\bar{x})$是$\left{f\left(x_k\right)\right}$的极限点，所以整个序列$\left{f\left(x_k\right)\right}$收敛到$f(\bar{x})$。因此， $$ f\left(x_k\right)-f\left(x{k+1}\right) \rightarrow 0
$$
根据Armijo法则的定义和方向$d_k$的下降性质$\nabla f\left(x_k\right)^{\prime} d_k \leq$ 0，我们有
$$
f\left(x_k\right)-f\left(x_{k+1}\right) \geq-\sigma \alpha_k \nabla f\left(x_k\right)^{\prime} d_k \geq 0,
$$
通过结合前面两个关系，
$$
\alpha_k \nabla f\left(x_k\right)^{\prime} d_k \rightarrow 0
$$
从方程(2.72)的左侧和假设$\nabla f(\bar{x}) \neq 0$可以得出如下结论
$$
\limsup {\substack{k \rightarrow \infty \ k \in \mathcal{K}}} \nabla f\left(x_k\right)^{\prime} d_k<0 $$与式(2.73)一起，意味着$$ \left{\alpha_k\right}{\mathcal{K}} \rightarrow 0 . $$ 它与式(2.73)一起意味着 $$ \left{\alpha_k\right}_{\mathcal{K}} \rightarrow 0 $$ 由于$\alpha_k$的初始试验值$s_k$与$0, s_k$有界限，因此对于所有大于某个迭代索引$\bar{k}$的$k \in \mathcal{K}$，至少会减少一次。因此，我们必须为所有$k \in \mathcal{K}$ with $k>\bar{k}$
$$
f\left(x_k\right)-f\left(x_k+\left(\alpha_k / \beta\right) d_k\right)<-\sigma\left(\alpha_k / \beta\right) \nabla f\left(x_k\right)^{\prime} d_k
$$

数学代写|凸优化作业代写Convex Optimization代考|Convergence of Steepest Descent to a Single Limit

设$f: \Re^n \mapsto \Re$是一个可微凸函数，对于$L>0$，我们有
$$
|\nabla f(x)-\nabla f(y)| \leq L|x-y|, \quad \forall x, y \in \Re^n .
$$
设$X^$为$f$的最小值集合，并设$X^$为非空。考虑最陡下降法
$$
x_{k+1}=x_k-\alpha_k \nabla f\left(x_k\right)
$$
证明在以下两种步长规则条件下，$\left{x_k\right}$收敛于$f$的一个极小点:
(i)对于一些$\epsilon>0$，我们有
$$
\epsilon \leq \alpha_k \leq \frac{2(1-\epsilon)}{L}, \quad \forall k .
$$
(ii) $\alpha_k \rightarrow 0$和$\sum_{k=0}^{\infty} \alpha_k=\infty$。
注:原始源代码[BGI95]也显示了Armijo规则的一个变体收敛到单个极限。这应该与[Gon00]的结果形成对比，该结果表明，具有精确的线最小化规则的最陡下降方法可能产生具有多个极限点的序列(当然所有这些都是最优的)，即使对于凸代价函数也是如此。还有一个“局部捕获”定理，适用于非凸连续可微代价函数$f$的梯度方法和孤立的局部最小值$f$(在$x^$的邻域内唯一的局部最小值$x^$)。在温和的条件下，它断言存在一个以$x^$为中心的开放球体$S_{x^}$，这样一旦生成的序列$\left{x_k\right}$进入$S_{x^}$，它就收敛到$x^$(参见[Ber82a]， Prop. 1.12，或[Ber99]， Prop. 1.2.5以及那里给出的参考文献)。简短证明:考虑步长规则(i)。从下降不等式(练习2.2)中，我们得到所有$k$
$$
f\left(x_{k+1}\right) \leq f\left(x_k\right)-\alpha_k\left(1-\frac{\alpha_k L}{2}\right)\left|\nabla f\left(x_k\right)\right|^2 \leq f\left(x_k\right)-\epsilon^2\left|\nabla f\left(x_k\right)\right|^2
$$
所以$\left{f\left(x_k\right)\right}$是单调不增加的并且是收敛的。将上述关系对所有的$k$值相加，取极限$k \rightarrow \infty$，则对所有的$x^* \in X^$， $$ f\left(x^\right) \leq f\left(x_0\right)-\epsilon^2 \sum_{k=0}^{\infty}\left|\nabla f\left(x_k\right)\right|^2
$$

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金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语 广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

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随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。 随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习和应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。