Is this quadratic form the Sharpe ratio?

Question

I'm reading Merton's An Analytic Derivation of the Efficient Portfolio Frontier. In section IV, he derives the efficient frontier with a riskless asset. Let $\mathbf{w}$ be a vector of portfolio weights and let $w_f$ be the weight of the risk-free asset $r_f$. Then

$$ \mathbf{w}^{\top} \mathbf{1} + w_f = 1 \tag{1} $$

by construction. The optimization problem is

$$ \begin{aligned} \min_{\mathbf{w}} &&& \mathbf{w}^{\top} \boldsymbol{\Sigma} \mathbf{w}, \\ \text{subject to} &&& \mathbf{w}^{\top} \tilde{\boldsymbol{\mu}} = \tilde{\mu}_p, \end{aligned} \tag{2} $$

where

$$ \begin{aligned} \tilde{\boldsymbol{\mu}} &\triangleq \boldsymbol{\mu} - r_f \mathbf{1}, \\ \tilde{\mu}_p &\triangleq \mu_p - r_f, \end{aligned} \tag{3} $$

and where $\boldsymbol{\mu}$ is a vector of expected returns and $\mu_p$ is the portfolio's return. I can write down the Lagrangian function and derive the first-order conditions:

$$ \begin{aligned} \nabla_{\mathbf{w}} \mathcal{L} &= 2 \boldsymbol{\Sigma} \mathbf{w} + \lambda \tilde{\boldsymbol{\mu}} = \mathbf{0}, \\ \frac{\partial}{\partial \lambda} \mathcal{L} &= \mathbf{w}^{\top} \tilde{\boldsymbol{\mu}} - \tilde{\mu}_p = 0. \end{aligned} \tag{4} $$

Finally, I can derive the same optimal weights

$$ \mathbf{w} = \tilde{\mu}_p \left( \frac{\boldsymbol{\Sigma}^{-1} \tilde{\boldsymbol{\mu}}}{\tilde{\boldsymbol{\mu}}^{\top} \boldsymbol{\Sigma}^{-1} \tilde{\boldsymbol{\mu}}} \right) \tag{5} $$

and the same quadratic-form equation as Merton, his equation 35:

$$ \begin{aligned} | \mu_p - r_f | &= \sigma_p \sqrt{(\boldsymbol{\mu} - r_f \mathbf{1})^{\top} \boldsymbol{\Sigma}^{-1} (\boldsymbol{\mu} - r_f \mathbf{1})} \\ &\Downarrow \\ \mu_p &= r_f \pm \sigma_p \sqrt{(\boldsymbol{\mu} - r_f \mathbf{1})^{\top} \boldsymbol{\Sigma}^{-1} (\boldsymbol{\mu} - r_f \mathbf{1})}. \end{aligned} \tag{6} $$

This is clearly a piecewise function in which each half is a linear function. The top half is, I assume, what people call the capital market line, since the independent variable is $\mu_p$, the dependent variable is $\sigma_p$, and the $y$-intercept is $r_f$. However, and this is my question, the slope is not the Sharpe ratio:

$$ \frac{\mu_p - r_f}{\sigma_p} \stackrel{???}{\neq} \sqrt{(\boldsymbol{\mu} - r_f \mathbf{1})^{\top} \boldsymbol{\Sigma}^{-1} (\boldsymbol{\mu} - r_f \mathbf{1})}. \tag{7} $$

What am I missing?

Answer 1

Perhaps this is helpful. Look at my answer to a related question to follow my notation better.

$$ \begin{align*}a &\equiv \sum_i \sum_j V_{ij} \mu_i \quad \quad \text{(in Merton paper)}\\ &= \boldsymbol{1}'V \boldsymbol{\mu} \quad \quad \text{(in vector notation)} \\ &= s_{1u} \quad\quad \text{my preferred shorthand} \\ \\ b &\equiv \sum_i \sum_j V_{ij} \mu_j \mu_k\\ &= \boldsymbol{\mu}'V \boldsymbol{\mu} \quad \quad \text{(in vector notation)} \\ &= s_{uu} \quad\quad \text{my preferred shorthand}\\ \\ c &\equiv \sum_i \sum_j V_{ij} \\ &= \boldsymbol{1}'V \boldsymbol{1} \quad \quad \text{(in vector notation)}\\ &= s_{11} \quad\quad \text{my preferred shorthand} \end{align*} $$

$$ S = \begin{bmatrix} b & a \\ a & c \end{bmatrix} $$

$$ \mathbf{w}^* = \frac{\Sigma^{-1} \left( \boldsymbol{\mu} - r_f\right)}{\mathbf{1}' \Sigma^{-1} \left( \boldsymbol{\mu} - r_f \right)}$$

$$ \mathbf{w^*}'\boldsymbol{\mu} = \frac{s_{uu} - r_f s_{1u}}{s_{1u} - r_fs_{11}}$$

$$ \mathbf{w^*}'\boldsymbol{\mu} - r_f = \frac{s_{uu} - 2r_fs_{1u} + r_f^2s_{11}}{s_{1u} - r_f s_{11}}$$

$$ \begin{align*} \mathbf{w^*}'\Sigma \mathbf{w^*} &= \frac{ \left( \boldsymbol{\mu} - r_f \mathbf{1}\right)'\Sigma^{-1} \left( \boldsymbol{\mu} - r_f \mathbf{1}\right) }{\left(s_{1u} - r_fs_{11} \right)^2} \\ &= \frac{s_{uu} - 2r_f s_{1u} + r_f^2 s_{11} }{\left(s_{1u} - r_fs_{11} \right)^2} \end{align*} $$ Finally: $$ \frac{\mathbf{w^*}'\boldsymbol{\mu} - r_f}{\sqrt{\mathbf{w^*}'\Sigma \mathbf{w^*}} } = \sqrt{s_{uu} - 2r_f s_{1u} + r_f^2 s_{11}}$$

Which is the last line you have in (7).