Tangency portfolio with two additional constraints so that portfolio weights are unconstrained

Question

I know that the formula for determining the weights of the Tangency portfolio is given as $w_{tan}$ = $\frac{\Sigma \mu}{\iota^{\prime}\Sigma\mu }$, but I was wondering how to derive the weights in case we add the constraints that the weights should be larger than or equal to -1, and smaller than or equal to 1.

I was wondering whether there is a closed form solution available, and/or what the derivation looks like?

I guess the optimisation problem would look something like this: $$ \frac{w^{\prime}\mu}{\sqrt{w^{\prime}\Sigma w}} $$ s.t. $$ w^{\prime}\iota = 1 $$ $$ w_{i} \geq -1, \forall i = 1, \dots, N $$ $$ w_{i} \leq 1, \forall i = 1, \dots, N $$

Please correct me if this is the wrong representation of the problem.

Answer 1

Yes there are two ways to solve the tangency portfolio:

• closed-form analytical solution
• optimization problem (maximization of the Sharpe ratio)

The closed-form analytical solution you incorrectly wrote is actually

$$w_{\text{tan}} = \frac{\Sigma^{-1} \left(\mu - r_f \cdot \iota\right)}{\iota^{\prime}\Sigma^{-1}\left(\mu - r_f \cdot \iota\right)}$$

This is already the unconstrained maximum Sharpe ratio portfolio, where "unconstrained" means short-selling is allowed so that weights can be larger than or equal to $-1$ (in other words, less than $0$), and smaller than or equal to $1$. For the equivalent optimization problem we therefore don't need to impose additional constraints (besides the sum-to-1 rule) in order to "unconstrain" the weights since the unconstrained portfolio is an unconstrained problem.

On the other hand, the constrained max Sharpe portfolio, where short-sales are forbidden, does not have a simple closed-form analytical solution like the unconstrained portfolio, and does require an additional constraint of non-negativity, $w_i \geq 0 \enspace \forall i=1,\dots,N$.