# Is this quadratic form the Sharpe ratio?

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?

• The RHS is the Sharpe ratio of the Markowitz portfolio. Jan 5, 2022 at 17:42
• What's the "Markowitz portfolio"? Is this the same as the tangency portfolio? If that's the case, I'd be expect to be able to take the optimal weights for the tangency portfolio, plug them into the Sharpe ratio, and then get my equation 7. I don't see it yet, but maybe I'm missing some basic algebraic step.
– jds
Jan 5, 2022 at 18:27
• I'm not quite sure where your confusion is. The algebra works out. Also I'd call the top half the efficient frontier. People call it the capital market line in the CAPM model because in the CAPM model, the tangency portfolio is the market portfolio. In general, this is obviously not the case. Jan 5, 2022 at 18:56

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).