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I have recently implemented MPT in Python, however, when I allow negative weights (short selling), they do not add up to one. Isn't it suppose not to happen? On the other hand, when I don't allow, they add up to one as usual.

Here's what I did:

  1. Defined the return as usual: $\vec{W}^{T} \vec{R}$, where $\vec{W}$ have all the weights and $\vec{R}$ is the random vector of the returns;
  2. Calculated the expected returns and the covariance matrix all as percentages;
  3. Defined risk as the standard deviation, which satisfies: $\sigma^2(\vec{W}) = \vec{W}^{T} \Sigma \vec{W}$;
  4. Minimized $\sigma^2(\vec{W})$ under i) $\vec{W}^{T} \vec{I}=1$, where $\vec{I} = [1\,\ 1 \cdots 1]^{T}$ and ii) $\vec{W}^{T} \mathbb{E}[\vec{R}] = \mu_0$, where $\mu_0$ is given;
  5. Set the lagrangian as follows: $\mathcal{L}(\vec{W}, \lambda_1, \lambda_2) = \sigma^2(\vec{W}) + \lambda_1 (\vec{W}^{T} \vec{I} - 1) + \lambda_2 (\vec{W}^{T} \mathbb{E}[\vec{R}] - \mu_0)$
  6. Take all partial derivatives and end up in a system of equations like so:

\begin{equation} \begin{bmatrix} 2 \Sigma \,\,\, \vec{I} \,\,\, \mathbb{E}[\vec{R}] \\ \vec{I}^{T} \,\,\, 0 \,\,\, 0 \\ \mathbb{E}[\vec{R}]^{T} \,\,\, 0 \,\,\, 0 \end{bmatrix} \cdot \begin{bmatrix} \vec{W} \\ \lambda_1 \\ \lambda_2 \end{bmatrix} = \begin{bmatrix} \vec{0} \\ 1 \\ \mu_0 \end{bmatrix} \end{equation}

  1. Since this is a system of $A \vec{x} = \vec{b}(\mu_0)$ kind and $\vec{b}$ is parametrized by $\mu_0$, I used QR factorization to solve it for every $\mu_0$;

  2. Once I have the solution $\vec{x}$, I have the weights given by $\vec{W}$ as well. Therefore, it's been solved.

Is there anything missing? Does it happen commonly?

Thanks

EDIT

Thank you guys, I managed to solve the problem. The function I created to calculate all the weights was incorrect (I was using norm 1).

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  • $\begingroup$ The second row of the equation in 6 says, in English "the sum of the elements in W is equal to 1". So how can it be that in the "solution" the elements of W do not add up to 1? It can only happen if the system is not solved properly. $\endgroup$
    – nbbo2
    Oct 20, 2021 at 14:26
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    $\begingroup$ Your problem is solved by $\vec{W} = \Sigma^{-1}\left(c_1 \vec{I} + c_2 E\left[\vec{R}\right]\right)$. Just use the two equality conditions to solve for $c_i$. This is easy to do analytically because you have no positivity constraint. $\endgroup$
    – shabbychef
    Oct 20, 2021 at 16:26
  • $\begingroup$ @noob2 Indeed. Then all steps are correct I suppose. Thanks. I thought my QR factorization was working just fine.... To calculate the inverse of the matrix R, I followed Lewis' answer: math.stackexchange.com/questions/1003801/… . Maybe the problem is there. $\endgroup$
    – Mr. N
    Oct 20, 2021 at 18:00
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    $\begingroup$ Try to work around inverting matrices although I doubt that this is your problem right now. It's easy to check whether your matrix inversion code is correct: multiply by the original and check that you get the identity back. $\endgroup$
    – Bob Jansen
    Oct 20, 2021 at 18:13
  • $\begingroup$ @BobJansen QR factorization looks correct. Thanks I'll check the inverses and work as the links says. $\endgroup$
    – Mr. N
    Oct 20, 2021 at 18:56

1 Answer 1

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We will use bold notation for vectors and write $\pmb \mu = \mathbb E[\vec R]$.


From the first equation you have $$\pmb w = \frac 12 \pmb\Sigma^{-1}\left(\pmb 1^T\lambda_1 + \pmb\mu\lambda_2\right).$$ Hence, we will first find $\lambda_1$ and $\lambda_2$. Using the constraints, we may write \begin{aligned} 1 = \pmb1^T\pmb w &= \frac 12 \pmb1^T\pmb\Sigma^{-1}\left(\pmb 1^T\lambda_1 + \pmb\mu\lambda_2\right)\\ \mu_0 = \pmb \mu^T\pmb w &= \frac 12 \pmb\mu^T\pmb\Sigma^{-1}\left(\pmb 1^T\lambda_1 + \pmb\mu\lambda_2\right) \end{aligned} In other words, we need to solve the system

$$ \frac 12\begin{bmatrix} \pmb 1^T \pmb \Sigma^{-1}\pmb 1^T & \pmb 1^T \pmb \Sigma^{-1}\pmb \mu^T\\ \pmb 1^T \pmb \Sigma^{-1}\pmb \mu^T & \pmb \mu^T \pmb \Sigma^{-1}\pmb \mu^T \end{bmatrix}\begin{bmatrix} \lambda_1\\ \lambda_2 \end{bmatrix} = \begin{bmatrix} 1\\ \mu_0 \end{bmatrix}. $$

The solution is given by $$ \pmb\lambda = \frac{2}{(\pmb 1^T \pmb \Sigma^{-1}\pmb 1^T)(\pmb \mu^T \pmb \Sigma^{-1}\pmb \mu^T) - (\pmb 1^T \pmb \Sigma^{-1}\pmb \mu^T)^2}\begin{bmatrix} \pmb \mu^T \pmb \Sigma^{-1}\pmb \mu^T & -\pmb 1^T \pmb \Sigma^{-1}\pmb \mu^T\\ -\pmb 1^T \pmb \Sigma^{-1}\pmb \mu^T & \pmb 1^T \pmb \Sigma^{-1}\pmb 1^T \end{bmatrix}\begin{bmatrix} 1\\ \mu_0 \end{bmatrix}. $$

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