## 金融代写|金融数值计算代写Market Risk, Numerical Analysis for Finance代考|Lagrange multipliers

In applications, it is often necessary to optimize functions under some constraints. The Lagrange multipliers theorem $3.22$ provides necessary optimum conditions for a problem of the following kind:
$\max f(\boldsymbol{x}) \quad$ subject to $g(\boldsymbol{x})=0$
or $\min f(\boldsymbol{x}) \quad$ subject to $g(\boldsymbol{x})=0$.
Theorem $3.22$ (Lagrange multipliers – general case). Let $m<n$, let $V$ be open in $\mathbb{R}^n$, and let $f, g_j: V \rightarrow \mathbb{R}$ be $\mathcal{C}^1$ on $V$, for $j=1,2 \ldots, m$. Suppose that:
$$\frac{\partial\left(g_1, \ldots, g_m\right)}{\partial\left(x_1, \ldots, x_n\right)}$$ has rank $m$ at $\boldsymbol{x}0 \in V$, where $g_j\left(\boldsymbol{x}_0\right)=0$ for $j=1,2, \ldots, m$. Assume further that $x_0$ is a local extremum for $f$ in the set: $$M=\left{\boldsymbol{x} \in V \quad \mid \quad g_j(\boldsymbol{x})=0\right} .$$ Then, there exist scalars $\lambda_1, \ldots, \lambda_m$, such that: $$\nabla\left(f\left(\boldsymbol{x}_0\right)-\sum{k=1}^m \lambda_k g_k\left(\boldsymbol{x}_0\right)\right)=\mathbf{0}$$
We will limit the proof of the Lagrange multipliers theorem $3.22$ in a twodimensional context. To this aim, it is first necessary to consider some preliminary results; we will resume the proof in $\S 3.8$.

## 金融代写|金融数值计算代写Market Risk, Numerical Analysis for Finance代考|Mean-Value theorem

We begin with recalling the definition of a segment in the Euclidean space.
Definition 3.23. Given $\boldsymbol{x}, \boldsymbol{y} \in \mathbb{R}^n$, the segment joining $\boldsymbol{x}$ and $\boldsymbol{y}$ is defined as:
$$[\boldsymbol{x}, \boldsymbol{y}]:=\left{\boldsymbol{z} \in \mathbb{R}^n \quad \mid \boldsymbol{z}=t \boldsymbol{x}+(1-t) \boldsymbol{y}, \quad 0 \leq t \leq 1\right}$$
The one-dimensional Mean-Value theorem (already met in Example 3.3), also called Lagrange Mean-Value theorem or First Mean-Value theorem, can be extended to the Euclidean space $\mathbb{K}^n$.

Theorem $3.24$ (Mean-Value). Let $A \subset \mathbb{R}^n$, and $f: A \rightarrow \mathbb{R}$. Consider $\boldsymbol{x}, \boldsymbol{y} \in \mathbb{R}^n$ such that $[\boldsymbol{x}, \boldsymbol{y}] \subset A^{\circ}$, the interior of $A$ (see Definition 1.18). Assume that $f(\boldsymbol{x})$ is differentiable in $[\boldsymbol{x}, \boldsymbol{y}]$. Then, there exists $\boldsymbol{z} \in[\boldsymbol{x}, \boldsymbol{y}]$ such that:
$$f(\boldsymbol{x})-f(\boldsymbol{y})=\nabla f(\boldsymbol{z}) \cdot(\boldsymbol{x}-\boldsymbol{y})$$
Proof. Define $\varphi:[0,1] \rightarrow \mathbb{R}^n, \varphi(t)=\boldsymbol{y}+t(\boldsymbol{x}-\boldsymbol{y})$. Observe that $\varphi \in \mathcal{C}$ and it is differentiable for any $t \in(0,1)$. Moreover, $\varphi^{\prime}(t)=\boldsymbol{x}-\boldsymbol{y}$. It follows that $g=f \circ \varphi:[0,1] \rightarrow \mathbb{R}$ is continuous and differentiable in $(0,1)$. We can thus apply the one-dimensional version of the Mean-Value theorem, to infer the existence of $\eta \in(0,1)$ such that:
$$f(\boldsymbol{x})-f(\boldsymbol{y})=g(1)-g(0)=g^{\prime}(\eta) .$$
On the other hand, the Chain Rule implies:
$$g^{\prime}(\eta)=\nabla f(\varphi(\eta)) \cdot \varphi^{\prime}(\eta)=\nabla f(\varphi(\eta)) \cdot(\boldsymbol{x}-\boldsymbol{y}) .$$
Since $\boldsymbol{z}=\varphi(\eta) \in[\boldsymbol{x}, \boldsymbol{y}]$, Theorem $3.24$ is proved.

# 金融数值计算代考

## 金融代写|金融数值计算代写市场风险，金融的数值分析代考|拉格朗日乘数

$\max f(\boldsymbol{x}) \quad$受制于$g(\boldsymbol{x})=0$

$$\frac{\partial\left(g_1, \ldots, g_m\right)}{\partial\left(x_1, \ldots, x_n\right)}$$在$\boldsymbol{x}0 \in V$上的排名为$m$，其中$g_j\left(\boldsymbol{x}_0\right)=0$为$j=1,2, \ldots, m$。进一步假设$x_0$是$f$在集合中的局部极值:$$M=\left{\boldsymbol{x} \in V \quad \mid \quad g_j(\boldsymbol{x})=0\right} .$$那么，存在标量$\lambda_1, \ldots, \lambda_m$，这样:$$\nabla\left(f\left(\boldsymbol{x}_0\right)-\sum{k=1}^m \lambda_k g_k\left(\boldsymbol{x}_0\right)\right)=\mathbf{0}$$

## 金融代写|金融数值计算代写市场风险，金融数值分析代考|中值定理

$$[\boldsymbol{x}, \boldsymbol{y}]:=\left{\boldsymbol{z} \in \mathbb{R}^n \quad \mid \boldsymbol{z}=t \boldsymbol{x}+(1-t) \boldsymbol{y}, \quad 0 \leq t \leq 1\right}$$

$$f(\boldsymbol{x})-f(\boldsymbol{y})=\nabla f(\boldsymbol{z}) \cdot(\boldsymbol{x}-\boldsymbol{y})$$

$$f(\boldsymbol{x})-f(\boldsymbol{y})=g(1)-g(0)=g^{\prime}(\eta) .$$

$$g^{\prime}(\eta)=\nabla f(\varphi(\eta)) \cdot \varphi^{\prime}(\eta)=\nabla f(\varphi(\eta)) \cdot(\boldsymbol{x}-\boldsymbol{y}) .$$

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