# 金融代写|风险和利率理论代写Market Risk, Measures and Portfolio Theory代考|GBUS846

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

In this section we give the formula for the weights of the portfolio with smallest variance. Before doing so, we need to consider a technical lemma.
Lemma $4.3$
We have the following formulae for the gradients computed with respect to w:
\begin{aligned} \nabla\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right) &=\boldsymbol{\mu}, \ \nabla\left(\mathbf{w}^{\mathrm{T}} \mathbf{1}\right) &=\mathbf{1}, \ \nabla\left(\mathbf{w}^{\mathrm{T}} C \mathbf{w}\right) &=2 C \mathbf{w}, \end{aligned}
and the Hessian of $\mathbf{w}^{\mathrm{T}} C \mathbf{w}$ is equal to $2 C$.
Proof Since
$$\frac{\partial}{\partial w_i}\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right)=\frac{\partial}{\partial w_i}\left(w_1 \mu_1+\cdots+w_n \mu_n\right)=\mu_i$$

we see that
$$\nabla\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right)=\left[\begin{array}{c} \frac{\partial}{\partial w_1}\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right) \ \vdots \ \frac{\partial}{\partial w_n}\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right) \end{array}\right]=\left[\begin{array}{c} \mu_1 \ \vdots \ \mu_n \end{array}\right]=\boldsymbol{\mu},$$
which proves (4.3).
The proof of (4.4) follows from an identical argument, using 1 instead of $\mu$.
To prove (4.5) we observe that in
$$\frac{\partial}{\partial w_i}\left(\mathbf{w}^{\mathrm{T}} C \mathbf{w}\right)=\frac{\partial}{\partial w_i} \sum_{j=1}^n \sum_{k=1}^n w_j w_k \sigma_{j k}$$
the derivative of each term can be non-zero only when $j=i$ or $k=i$. This means that
\begin{aligned} \frac{\partial}{\partial w_i} & \sum_{j=1}^n \sum_{k=1}^n w_j w_k \sigma_{j k} \ &=\frac{\partial}{\partial w_i}\left(w_i w_i \sigma_{i i}+\sum_{j=i} \sum_{k \neq i} w_j w_k \sigma_{j k}+\sum_{j \neq i} \sum_{k=i} w_j w_k \sigma_{j k}\right) \ &=2 w_i \sigma_{i i}+\sum_{k \neq i} w_k \sigma_{i k}+\sum_{j \neq i} w_j \sigma_{j i} \ &\left.=2 \sum_{k=1}^n w_k \sigma_{i k} \quad \quad \text { (since } \sigma_{j i}=\sigma_{i j}\right) \ &=2(C \mathbf{w})_i \end{aligned}
where $(C \mathbf{w})_i$ stands for the $i$-th coordinate of the vector $C \mathbf{w}$. Combining the partial derivatives on all coordinates gives (4.5).

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

To find the efficient frontier, we have to recognise and eliminate the dominated portfolios. To this end we fix a level of expected return, denote it by $m$, and consider all portfolios with $\mu_{\mathrm{w}}=m$. All of these are redundant except the one with the smallest variance. The family of such portfolios, parameterised by $m$, is called the minimum variance line (see Figure 4.6).
More precisely, portfolios on the minimum variance line are solutions of the following problem:
$$\begin{array}{ll} \min \mathbf{w}^{\mathrm{T}} C \mathbf{w}, & \ \text { subject to: } & \mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}=m, \ & \mathbf{w}^{\mathrm{T}} \mathbf{1}=1 . \end{array}$$

Let $M$ be $a 2 \times 2$ matrix of the form
$$M=\left[\begin{array}{cc} \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \boldsymbol{\mu} & \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1} \ \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1} & \mathbf{1}^{\mathrm{T}} C^{-1} \mathbf{1} \end{array}\right] .$$
If $C$ and $M$ are invertible, then the solution of problem (4.12) is given by
$$\mathbf{w}=\frac{1}{\operatorname{det}(M)} C^{-1}\left(\operatorname{det}\left(M_1\right) \mu+\operatorname{det}\left(M_2\right) \mathbf{1}\right),$$
where
$$M_1=\left[\begin{array}{cc} m & \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1} \ 1 & \mathbf{1}^{\mathrm{T}} C^{-1} \mathbf{1} \end{array}\right], \quad M_2=\left[\begin{array}{cc} \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \boldsymbol{\mu} & m \ \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1} & 1 \end{array}\right] .$$
Proof We introduce the Lagrange multiplier $\lambda=\left(\lambda_1, \lambda_2\right)$, and the Lagrangian
$$L(\mathbf{w})=\nabla\left(\mathbf{w}^{\mathrm{T}} C \mathbf{w}\right)-\lambda_1 \nabla\left(\mathbf{w}^{\mathrm{T}} \mu-m\right)+\lambda_2 \nabla\left(\mathbf{w}^{\mathrm{T}} \mathbf{1}-1\right)=0 .$$
Using Lemma $4.3$ we can compute
$$L(\mathbf{w})=2 C \mathbf{w}-\lambda_1 \boldsymbol{\mu}-\lambda_2 \mathbf{1}=0 .$$
We solve this system for $\mathbf{w}$ :
$$\mathbf{w}=\frac{1}{2} \lambda_1 C^{-1} \boldsymbol{\mu}+\frac{1}{2} \lambda_2 C^{-1} \mathbf{1} .$$
Since $\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}=\boldsymbol{\mu}^{\mathrm{T}} \mathbf{w}$ and $\mathbf{w}^{\mathrm{T}} \mathbf{1}=\mathbf{1}^{\mathrm{T}} \mathbf{w}$, substituting (4.14) into the constraints from (4.12), we obtain a system of linear equations
\begin{aligned} &\frac{1}{2} \lambda_1 \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \boldsymbol{\mu}+\frac{1}{2} \lambda_2 \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1}=m, \ &\frac{1}{2} \lambda_1 \mathbf{1}^{\mathrm{T}} C^{-1} \boldsymbol{\mu}+\frac{1}{2} \lambda_2 \mathbf{1}^{\mathrm{T}} C^{-1} \mathbf{1}=1 . \end{aligned}

# 风险和利率理论代考

## 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|最小方差投资组合

\begin{aligned} \nabla\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right) &=\boldsymbol{\mu}, \ \nabla\left(\mathbf{w}^{\mathrm{T}} \mathbf{1}\right) &=\mathbf{1}, \ \nabla\left(\mathbf{w}^{\mathrm{T}} C \mathbf{w}\right) &=2 C \mathbf{w}, \end{aligned}

$$\frac{\partial}{\partial w_i}\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right)=\frac{\partial}{\partial w_i}\left(w_1 \mu_1+\cdots+w_n \mu_n\right)=\mu_i$$

$$\nabla\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right)=\left[\begin{array}{c} \frac{\partial}{\partial w_1}\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right) \ \vdots \ \frac{\partial}{\partial w_n}\left(\mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}\right) \end{array}\right]=\left[\begin{array}{c} \mu_1 \ \vdots \ \mu_n \end{array}\right]=\boldsymbol{\mu},$$
，证明(4.3)。
(4.4)的证明来自一个相同的论证，用1代替 $\mu$.

$$\frac{\partial}{\partial w_i}\left(\mathbf{w}^{\mathrm{T}} C \mathbf{w}\right)=\frac{\partial}{\partial w_i} \sum_{j=1}^n \sum_{k=1}^n w_j w_k \sigma_{j k}$$
，只有当 $j=i$ 或 $k=i$。这意味着
\begin{aligned} \frac{\partial}{\partial w_i} & \sum_{j=1}^n \sum_{k=1}^n w_j w_k \sigma_{j k} \ &=\frac{\partial}{\partial w_i}\left(w_i w_i \sigma_{i i}+\sum_{j=i} \sum_{k \neq i} w_j w_k \sigma_{j k}+\sum_{j \neq i} \sum_{k=i} w_j w_k \sigma_{j k}\right) \ &=2 w_i \sigma_{i i}+\sum_{k \neq i} w_k \sigma_{i k}+\sum_{j \neq i} w_j \sigma_{j i} \ &\left.=2 \sum_{k=1}^n w_k \sigma_{i k} \quad \quad \text { (since } \sigma_{j i}=\sigma_{i j}\right) \ &=2(C \mathbf{w})_i \end{aligned}
where $(C \mathbf{w})_i$ 代表 $i$向量的-th坐标 $C \mathbf{w}$。将所有坐标上的偏导数组合得到(4.5).

## 金融代写|风险和利率理论代写市场风险、度量和投资组合理论代考|最小方差线

$$\begin{array}{ll} \min \mathbf{w}^{\mathrm{T}} C \mathbf{w}, & \ \text { subject to: } & \mathbf{w}^{\mathrm{T}} \boldsymbol{\mu}=m, \ & \mathbf{w}^{\mathrm{T}} \mathbf{1}=1 . \end{array}$$

$$M=\left[\begin{array}{cc} \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \boldsymbol{\mu} & \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1} \ \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1} & \mathbf{1}^{\mathrm{T}} C^{-1} \mathbf{1} \end{array}\right] .$$

$$\mathbf{w}=\frac{1}{\operatorname{det}(M)} C^{-1}\left(\operatorname{det}\left(M_1\right) \mu+\operatorname{det}\left(M_2\right) \mathbf{1}\right),$$
where
$$M_1=\left[\begin{array}{cc} m & \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1} \ 1 & \mathbf{1}^{\mathrm{T}} C^{-1} \mathbf{1} \end{array}\right], \quad M_2=\left[\begin{array}{cc} \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \boldsymbol{\mu} & m \ \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1} & 1 \end{array}\right] .$$我们引入拉格朗日乘子 $\lambda=\left(\lambda_1, \lambda_2\right)$，拉格朗日量
$$L(\mathbf{w})=\nabla\left(\mathbf{w}^{\mathrm{T}} C \mathbf{w}\right)-\lambda_1 \nabla\left(\mathbf{w}^{\mathrm{T}} \mu-m\right)+\lambda_2 \nabla\left(\mathbf{w}^{\mathrm{T}} \mathbf{1}-1\right)=0 .$$

$$L(\mathbf{w})=2 C \mathbf{w}-\lambda_1 \boldsymbol{\mu}-\lambda_2 \mathbf{1}=0 .$$我们解这个方程组 $\mathbf{w}$ :
$$\mathbf{w}=\frac{1}{2} \lambda_1 C^{-1} \boldsymbol{\mu}+\frac{1}{2} \lambda_2 C^{-1} \mathbf{1} .$$

\begin{aligned} &\frac{1}{2} \lambda_1 \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \boldsymbol{\mu}+\frac{1}{2} \lambda_2 \boldsymbol{\mu}^{\mathrm{T}} C^{-1} \mathbf{1}=m, \ &\frac{1}{2} \lambda_1 \mathbf{1}^{\mathrm{T}} C^{-1} \boldsymbol{\mu}+\frac{1}{2} \lambda_2 \mathbf{1}^{\mathrm{T}} C^{-1} \mathbf{1}=1 . \end{aligned}

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