(fin-edu:investing:mpt-capm:analytical)= # MPT and CAPM: analytical solution ## MPT Analytical solution of the MPT optimization problem for a **fully invested portfolio** with **no risk-free asset** $$ \mathbf{w}^* = \text{argmin}_{\mathbf{w}} \sigma^2 \quad \text{s.t.} \quad \begin{aligned} & \mathbf{w}^T \boldsymbol\mu = \overline{\mu} \\ & \sum_{i=1}^N w_i = 1 \end{aligned} $$ without any additional constraint. Using Lagrange multiplier method, the augmented objective function $$\widetilde{J}(\mathbf{w}; a, b) = \frac{1}{2} \mathbf{w}^T \boldsymbol\sigma^2 \mathbf{w} - a ( \mathbf{w}^T \boldsymbol\mu - \mu ) - b ( \mathbf{w}^T \mathbf{1} - 1 )$$ and setting its gradient equal to zero gives the following system of equations $$\begin{cases} \mathbf{0} & = \boldsymbol\sigma^2 \mathbf{w}^* - a \boldsymbol\mu - b\mathbf{1} \\ 0 & = \boldsymbol\mu^T \mathbf{w}^* - \mu \\ 0 & = \mathbf{1}^T \mathbf{w}^* - 1 \ , \end{cases}$$ or using matrix formalism $$\begin{bmatrix} \boldsymbol\sigma^2 & -\boldsymbol{\mu} & - \mathbf{1} \\ - \boldsymbol\mu^T & 0 & 0 \\ -\mathbf{1}^T & 0 & 0 \end{bmatrix} \begin{bmatrix} \mathbf{w}^* \\ a \\ b \end{bmatrix} = \begin{bmatrix} \mathbf{0} \\ -\mu \\ -1 \end{bmatrix}$$ Without any risk-free asset, the covariance matrix is non-singular, and thus invertible. (Formally) solving the first equation for $\mathbf{w}$, $$\mathbf{w} = \boldsymbol\sigma^{-2} \boldsymbol\mu a + \boldsymbol\sigma^{-2} \mathbf{1} b \ ,$$ a system of 2 equations in 2 unknowns $a$, $b$ reads $$\begin{bmatrix} \boldsymbol{\mu}^T \boldsymbol\sigma^{-2} \boldsymbol{\mu} & \boldsymbol{\mu}^T \boldsymbol\sigma^{-2} \mathbf{1} \\ \mathbf{1 }^T \boldsymbol\sigma^{-2} \boldsymbol{\mu} & \mathbf{1}^T \boldsymbol\sigma^{-2} \mathbf{1} \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \mu \\ 1 \end{bmatrix} \ .$$ and thus $$\begin{aligned} \begin{bmatrix} a \\ b \end{bmatrix} & = \frac{1}{A_{11}A_{22} - 2 A_{12}} \begin{bmatrix} A_{22} & -A_{12} \\ - A_{12} & A_{11} \end{bmatrix} \begin{bmatrix} \mu \\ 1 \end{bmatrix} \\ & = \frac{1}{\Delta} \left( \begin{bmatrix} A_{22} \\ -A_{12} \end{bmatrix} \mu + \begin{bmatrix} -A_{12} \\ A_{11} \end{bmatrix} \right) \ . \end{aligned}$$ Thus, the optimal asset allocation is a 1-degree function of $\mu$, $$\mathbf{w}^* = \mathbf{w}_0 + \mathbf{w}_{/\mu} \mu \ .$$ and its variance is a 2-degree function of $\mu$, $$\begin{aligned} \sigma^{2 *} & = \mathbf{w}^{* T} \boldsymbol\sigma^2 \mathbf{w}^* = \sigma^2_{0} + \sigma^2_1 \mu + \sigma^2_2 \mu^2 = \\ & = \left( a \boldsymbol\mu^T + b \mathbf{1}^T \right) \boldsymbol\sigma^{-2} \, \underbrace{ \boldsymbol\sigma^2 \, \boldsymbol\sigma^{-2} }_{ = \mathbf{I}} \left( \boldsymbol\mu a + \mathbf{1} b \right) = \\ & = a^2 \boldsymbol\mu^T \boldsymbol\sigma^{-2} \boldsymbol\mu + 2 a b \mathbf{1}^T \boldsymbol\sigma^2 \boldsymbol\mu + b^2 \mathbf{1}^T \boldsymbol\sigma^{-2} \mathbf{1} = \\ & = \left( \frac{1}{\Delta} \left( A_{22} \mu - A_{12} \right) \right)^2 A_{11} + 2 \frac{1}{\Delta^2} \left( A_{22}\mu - A_{12} \right) \frac{1}{\Delta} \left( -A_{12} \mu + A_{11} \right) A_{12} + \left( \frac{1}{\Delta} \left( -A_{12} \mu + A_{11} \right) \right)^2 A_{22} = \\ & = \frac{1}{\Delta^2} \left( \mu^2 \left( A_{22}^2 A_{11} - 2 A_{22} A_{12}^2 + A_{12}^2 A_{22} \right) + \mu \cdot 2 \left(- A_{11}A_{22}A_{12} + A_{22} A_{11} A_{12} + A_{12}^3 - A_{11}A_{22}A_{12} \right) + \left( A_{12}^2 A_{11} - 2 A_{12}^2 A_{11} + A_{11}^2 A_{22} \right) \right) = \\ & = \frac{1}{\Delta^2} \left[ \mu^2 A_{22} \left( A_{11} A_{22} - A_{12} \right) - 2 \mu A_{12} \left( A_{11} A_{22} - A_{12}^2 \right) + A_{11} \left( A_{11} A_{22} - A_{12}^2 \right) \right] = \\ & = \frac{1}{\Delta} \left( \mu^2 A_{22} - 2 \mu A_{12} + A_{11} \right) = \\ & = \begin{bmatrix} \mu & 1 \end{bmatrix} \mathbf{A}^{-1} \begin{bmatrix} \mu \\ 1 \end{bmatrix} \ . \end{aligned}$$ As the matrix $\mathbf{A}$ is definite positive (its inverse is definite positive as well?), it follows that $\sigma^2 > 0$ for any value of $\mu$, as expected for the value of a variance. **Some analytic geometry.** The function $$\begin{aligned} \text{Var}[r](\mu) & = \frac{1}{\Delta} \left( A_{22} \mu^2 - 2 A_{12} \mu + A_{11} \right) = \\ & = \frac{A_{22}}{\Delta} \left( \mu - \frac{A_{12}}{A_{22}} \right)^2 + \frac{A_{11}}{\Delta} - \frac{A_{12}^2}{A_{22} \, \Delta} = \\ & = \frac{A_{22}}{\Delta} \left( \mu - \frac{A_{12}}{A_{22}} \right)^2 + \frac{1}{A_{22}} \ . \end{aligned}$$ is the function of a parabola, with vertex in $$\begin{aligned} \mu_v & = \frac{A_{12}}{A_{22}} \\ \text{Var}[r]_v & = \text{Var}[r](\mu_v) = \frac{1}{\Delta} \left( \frac{-A_{12}^2 + A_{11}A_{22}}{A_{22}} \right) = \frac{1}{A_{22}} \ . \end{aligned}$$ Using $\sigma$ as an independent coordinate (and not $\text{Var}[r] = \sigma^2$)... ```{dropdown} Properties of matrix $\ \mathbf{A}$ Is it positive definite? Covariance matrix is positive matrix, so for $\forall \mathbf{v}$ $$ 0 < \mathbf{v} \boldsymbol\sigma^2 \mathbf{v} \ ,$$ and choosing $\mathbf{v} = \begin{bmatrix} \boldsymbol\mu & \mathbf{1} \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}$, for $\forall a, b$, it immediately follows $$\begin{aligned} 0 & < \mathbf{v} \boldsymbol\sigma^2 \mathbf{v} = \\ & = \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} \boldsymbol\mu^T \\ \mathbf{1}^T \end{bmatrix} \boldsymbol\sigma^2 \begin{bmatrix} \boldsymbol\mu & \mathbf{1} \end{bmatrix} \begin{bmatrix} a & b \end{bmatrix} = \\ & = \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} \boldsymbol\mu^T \boldsymbol\sigma^2 \boldsymbol\mu & \boldsymbol\mu^T \boldsymbol\sigma^2 \mathbf{1} \\ \mathbf{1}^T \boldsymbol\sigma^2 \boldsymbol\mu & \mathbf{1}^T \boldsymbol\sigma^2 \mathbf{1}\end{bmatrix} \begin{bmatrix} a & b \end{bmatrix} \ , \end{aligned}$$ and thus matrix $\mathbf{A}$ is definite positive. ``` ## CAPM Analytical solution of the MPT-CAPM optimization problem, with a risk-free asset. If a risk-free asset exists, the covariance matrix is singular. However, the risk-free asset can be partitioned from the risky assets, so that the covariance matrix of the return of the risky asset is non-singular. The problem becomes $$( \mathbf{w}, w_0 )^* = \text{argmin}_{()} \sigma^2 \quad \text{s.t} \quad \dots $$ ... $$ \begin{bmatrix} \boldsymbol\sigma^2 & \mathbf{0} & -\boldsymbol{\mu} & - \mathbf{1} \\ \mathbf{0}^T & 0 & -\mu_0 & -1 \\ - \boldsymbol\mu^T & -\mu_0 & 0 & 0 \\ -\mathbf{1}^T & -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} \mathbf{w}^* \\ w_0^* \\ a \\ b \end{bmatrix} = \begin{bmatrix} \mathbf{0} \\ 0 \\ -\mu \\ -1 \end{bmatrix} $$ From the second and the fourth equation, $$\begin{aligned} b & = - \mu_0 a \\ w^*_0 & = - \mathbf{1}^T \mathbf{w}^* + 1 \end{aligned}$$ and thus $$ \begin{bmatrix} \boldsymbol\sigma^2 & -\boldsymbol{\mu} + \mu_0 \mathbf{1} \\ - \boldsymbol\mu^T + \mu_0 \mathbf{1}^T & 0 \\ \end{bmatrix} \begin{bmatrix} \mathbf{w}^* \\ a \end{bmatrix} = \begin{bmatrix} \mathbf{0} \\ - \left( \mu - \mu_0 \right) \end{bmatrix} $$ whose solution reads $$\begin{aligned} a & = \frac{\mu_e}{\boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e} \\ \mathbf{w}^* & = \frac{\mu_e \boldsymbol\sigma^{-2} \boldsymbol\mu_e}{ \boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e } \end{aligned}$$ and the relationship between the variance and the expected value of the optimal portfolios, $$ \sigma^2 = \frac{\mu_e^2}{ \boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e } $$ or the linear relation between the standard deviation of the portoflio $\sigma$ and the excess return $\mu_e$ of the portfolio w.r.t. the risk-free asset, $$\sigma = \frac{\mu_e}{\sqrt{ \boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e }} \ .$$ ```{dropdown} Solution of the linear system - details $$\mathbf{w}^* = \boldsymbol\sigma^{-2} \left( \boldsymbol\mu - \mu_0 \mathbf{1} \right) a \ ,$$ $$\begin{aligned} \mu-\mu_0 & = \left( \boldsymbol\mu - \mu_0 \mathbf{1} \right)^T \mathbf{w}^* = \\ & = \left( \boldsymbol\mu - \mu_0 \mathbf{1} \right)^T \boldsymbol\sigma^{-2} \left( \boldsymbol\mu - \mu_0 \mathbf{1} \right) a \ , \end{aligned}$$ $$a = \frac{\mu_e}{\boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e} \ ,$$ and thus $$\mathbf{w}^* = \frac{\mu_e \boldsymbol\sigma^{-2} \boldsymbol\mu_e}{\boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e} \ .$$ Eventually, the variance of the portfolio reads $$\begin{aligned} \sigma^2 & = \mathbf{w}^{* \ T} \boldsymbol\sigma^{2} \mathbf{w} = \\ & = \mu_e^2 \dfrac{ \boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\sigma^{2} \boldsymbol\sigma^{-2} \boldsymbol\mu_e}{\left( \boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e \right)^2} = \\ & = \frac{\mu_e^2}{ \boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e } \ , \end{aligned}$$ with $\mu_e := \mu - \mu_0$ the excess desired return of the portfolio w.r.t. the risk-free asset, and $\boldsymbol\mu_e := \boldsymbol\mu - \mu_0 \mathbf{1}$ the vector of the excess returns of each risky asset w.r.t. the risk-free asset. Taking the square root of the last relation, a 1-degree function relates the standard deviation and the return of the portfolio, $$\sigma = \frac{\mu_e}{\sqrt{ \boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e }} \ .$$ ``` **Tangency condition** as a maximization of a measure of [risk-adjusted return](fin-edu:principles:rr), namely **Sharpe ratio** comparing the excess return and the variance of the portfolio w.r.t. a risk-free (zero variance) asset $(\cdot)_0$ used as a benchmark $(\cdot)_b$ $$S := \frac{\mathbb{E}[ r - r_b ]}{\sqrt{\text{Var}[ r - r_b ]}} = \frac{\mathbf{w}^{T} \boldsymbol\mu - \mu_0}{\sqrt{\mathbf{w}^{T} \boldsymbol\sigma^2 \mathbf{w}}}$$ as the variance reads $$\begin{aligned} \text{Var}[ r - r_0 ] & = \dots \\ \end{aligned}$$ ```{dropdown} Tangency condition between optimal portfolio lines w/ and w/o risk-free asset :open: W/o risk-free asset $$\sigma_{r}(\mu) = \sqrt{ \frac{A_{22} \mu^2 - 2 \mu A_{12} + A_{11}}{\Delta} } \ ,$$ with $A_{22} = \mathbf{1}^T \boldsymbol\sigma^{-2} \mathbf{1}$, $\mathbf{A}_{11} = \boldsymbol\mu^T \boldsymbol\sigma^{-2} \boldsymbol\mu$, $A_{12} = \boldsymbol\mu^T \boldsymbol\sigma^{-2} \mathbf{1}$, and $\Delta = A_{11} A_{22} - A_{12}$ W/ risk-free asset $$\sigma_f(\mu) = \frac{\mu - \mu_0}{\sqrt{ \left( \boldsymbol\mu - \mu_0 \mathbf{1} \right)^T \boldsymbol\sigma^{-2} \left( \boldsymbol\mu - \mu_0 \mathbf{1} \right) }} \ .$$ Tangency condition $$\begin{aligned} & \sigma_{r}(\overline{\mu}) = \sigma_{f}(\overline{\mu}) \\ & \sigma'_{r}(\overline{\mu}) = \sigma'_{f}(\overline{\mu}) \\ \end{aligned}$$ or with the variance, ... $$ 0 = \mu^2 \left( \frac{A_{22}}{\Delta} - \frac{1 }{\boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e} \right) -2\mu \left( \frac{A_{12}}{\Delta} - \frac{\mu_0 }{\boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e} \right) + \left( \frac{A_{11}}{\Delta} - \frac{\mu_0^2}{\boldsymbol\mu_e^T \boldsymbol\sigma^{-2} \boldsymbol\mu_e} \right) $$ ```