# 金融代写|风险和利率理论代写Market Risk, Measures and Portfolio Theory代考|FE630

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## 金融代写|风险和利率理论代写Market Risk, Measures and Portfolio Theory代考|Motivating examples

The aim of this section is to provide the underlying geometric intuition for the method.

We consider two functions
\begin{aligned} &f: \mathbb{R}^2 \rightarrow \mathbb{R}, \ &g: \mathbb{R}^2 \rightarrow \mathbb{R}, \end{aligned}
and show how to find solutions of the following problem:
Find
\begin{aligned} &\min f(x, y) \text {, } \ &\text { under the constraint: } g(x, y)=0 . \end{aligned}
Example 3.1
Consider
\begin{aligned} &f(x, y)=x^2+y^2, \ &g(x, y)=\frac{1}{2} x+\frac{1}{2} y-\frac{1}{2} . \end{aligned}
Basic arguments (say, by substituting $y=1-x$ into $f(x, y)$ and computing a derivative with respect to $x$ ) lead to the solution
$$x^=y^=\frac{1}{2} .$$
We now present an alternative approach. We first observe that one of the level curves $\left{(x, y): f(x, y)=r^2\right}$ (which are circles of radius $r$, as shown in Figure 3.1) is tangent at the point $\left(x^, y^\right)$ to the line ${(x, y): g(x, y)=0}$. Since the gradients
$$\begin{gathered} \nabla f(x, y)=\left[\begin{array}{c} \frac{\partial f}{\partial x}(x, y) \ \frac{\partial f}{\partial y}(x, y) \end{array}\right]=\left[\begin{array}{c} 2 x \ 2 y \end{array}\right], \ \nabla g(x, y)=\left[\begin{array}{c} \frac{\partial g}{\partial x}(x, y) \ \frac{\partial g}{\partial y}(x, y) \end{array}\right]=\left[\begin{array}{c} \frac{1}{2} \ \frac{1}{2} \end{array}\right], \end{gathered}$$
are orthogonal to the level curves, the vectors $\nabla f\left(x^, y^\right)$ and $\nabla g\left(x^, y^\right)$ should be collinear. This means that there should exist a number $\lambda \in \mathbb{R}$ such that we have the following system of two equations:
$$\nabla f(x, y)-\lambda \nabla g(x, y)=0$$

## 金融代写|风险和利率理论代写Market Risk, Measures and Portfolio Theory代考|Constrained extrema

The examples from the previous section have been considered on the plane. It turns out that a similar approach can be used in higher dimensions, and that we can consider more complicated constraints.

Our objective in this section is to show how to solve the following general constrained minimisation problem:
Find
$$\min f(\mathbf{v}) \text {, }$$
under the constraints: $\mathbf{g}(\mathbf{v})=\mathbf{0}$,
where
\begin{aligned} &f: \mathbb{R}^n \rightarrow \mathbb{R} \ &\mathbf{g}: \mathbb{R}^n \rightarrow \mathbb{R}^k \end{aligned}
We will provide necessary and, in the special case of quadratic forms, sufficient conditions for a solution to this problem.

To keep better track of dimensions, we use a bold font whenever we are dealing with vectors, and the normal font when dealing with numbers. Note that in stating the problem above we used $f$ for a function taking values in $\mathbb{R}$ and $\mathrm{g}$ for a function
$$\mathbf{g}(\mathbf{v})=\left(g_1(\mathbf{v}), \ldots, g_k(\mathbf{v})\right)$$
taking values in $\mathbb{R}^k$.
For the reader’s convenience we review some notations from multi-variable calculus.

# 风险和利率理论代写

## 金融代写|风险和利率理论代写市场风险、措施和投资组合理论代考|激励例子

\begin{aligned} &f: \mathbb{R}^2 \rightarrow \mathbb{R}, \ &g: \mathbb{R}^2 \rightarrow \mathbb{R}, \end{aligned}
，并展示如何找到以下问题的解决方案:
find
\begin{aligned} &\min f(x, y) \text {, } \ &\text { under the constraint: } g(x, y)=0 . \end{aligned}

\begin{aligned} &f(x, y)=x^2+y^2, \ &g(x, y)=\frac{1}{2} x+\frac{1}{2} y-\frac{1}{2} . \end{aligned}

$$x^=y^=\frac{1}{2} .$$

$$\begin{gathered} \nabla f(x, y)=\left[\begin{array}{c} \frac{\partial f}{\partial x}(x, y) \ \frac{\partial f}{\partial y}(x, y) \end{array}\right]=\left[\begin{array}{c} 2 x \ 2 y \end{array}\right], \ \nabla g(x, y)=\left[\begin{array}{c} \frac{\partial g}{\partial x}(x, y) \ \frac{\partial g}{\partial y}(x, y) \end{array}\right]=\left[\begin{array}{c} \frac{1}{2} \ \frac{1}{2} \end{array}\right], \end{gathered}$$

$$\nabla f(x, y)-\lambda \nabla g(x, y)=0$$

## 金融代写|风险和利率理论代写市场风险、措施和投资组合理论代考|约束极值

where
\begin{aligned} &f: \mathbb{R}^n \rightarrow \mathbb{R} \ &\mathbf{g}: \mathbb{R}^n \rightarrow \mathbb{R}^k \end{aligned}对于这个问题的解，我们将提供必要条件，并且在二次型的特殊情况下，提供充分条件

$$\mathbf{g}(\mathbf{v})=\left(g_1(\mathbf{v}), \ldots, g_k(\mathbf{v})\right)$$

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