The Quadratic Form for the Maximum Sharpe Ratio
Summary
The document examines the efficient frontier when a risk-free asset is available. It formulates minimum-variance portfolio selection for a target excess return and derives risky-asset weights using the inverse covariance matrix and expected excess returns. Substituting those weights into the frontier equation produces a quadratic form involving the excess-return vector and inverse covariance matrix. The question is whether the square root of this expression is the Sharpe ratio.
The included answer relates the quadratic form to the Sharpe ratio of the tangency portfolio: it represents the squared maximum Sharpe ratio, with its square root giving that ratio under the stated portfolio setup. The discussion uses alternative notation for covariance-based quantities and does not provide a full, clearly reconciled derivation of every displayed formula. Its result depends on the risk-free asset and the assumed expected returns and covariance matrix; estimated inputs can make the theoretical frontier differ from realized portfolio performance.
Key ideas
- With a risk-free asset, the efficient frontier can be described using portfolio excess returns and variance.
- The tangency portfolio weights depend on the inverse covariance matrix applied to expected excess returns.
- The quadratic form of excess returns and inverse covariance encodes the squared maximum Sharpe ratio.
- The square root of that quadratic form gives the maximum Sharpe ratio under the model assumptions.
- The result depends on expected return and covariance estimates, which may not match realized outcomes.
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Full text
# Is this quadratic form the Sharpe ratio?
# 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?
## Answer by Matthew Gunn (score 5, accepted)
https://quant.stackexchange.com/a/69367
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).Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.