I am trying to fit an auto-regressive model to a time-series where I have some constraints. We have the first order model, $$ X_{t+1} = AX_t + \xi_t, $$ which I can pose as a least-squares optimisation problem, $$ A^* = \arg \min_{A\in \Pi}|X_{1:T}-AX_{0:T-1}|^2 $$ but I have the constraints that, $A-I$ must be stable and that $I + \frac{1}{\tau}(A-I)$ must have non-negative entries. Can this still be solved numerically in R, Python or Matlab as a convex optimisation problem? The set of stable matrices is not convex but the set of non-negative matrices is convex.
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I will try make the model easier to solve:
- There is a set of given vectors $ \left\{ \boldsymbol{x}_{i} \in \mathbb{R}^{n} \right\}_{i = 1}^{T} $.
- Let $ \boldsymbol{X} = \begin{bmatrix} \boldsymbol{x}_{1} & \boldsymbol{x}_{2} & \cdots & \boldsymbol{x}_{T - 1} \end{bmatrix} $ and $ \boldsymbol{Y} = \begin{bmatrix} \boldsymbol{x}_{2} & \boldsymbol{x}_{3} & \cdots & \boldsymbol{x}_{T} \end{bmatrix} $.
Then the problem is given by:
$$ $$\begin{align} \arg \min_{\boldsymbol{A}} \quad & \frac{1}{2} {\left\| \boldsymbol{A} \boldsymbol{X} - \boldsymbol{Y} \right\|}_{F}^{2} \\ \text{subject to} \quad & \begin{aligned} \boldsymbol{A} - \boldsymbol{I} & \in \mathcal{S}_{+}^{n} \\ \boldsymbol{A} \phantom{ - \boldsymbol{I} \boldsymbol{I}} & \geq \boldsymbol{I} - \tau \boldsymbol{I} \end{aligned} \end{align}$$ $$
Where ${\left\| \cdot \right\|}_{F}$ is the Frobenius Norm and $\mathcal{S}_{+}^{n}$ is the set of Symmetric Positive Semi Definite (SPSD) matrices of size $n \times n$.
The projection onto the 2 constraints is given:
- Variation of Least Squares with Symmetric Positive Semi Definite (PSD) Constraint.
- Projection onto a Box Constraints.
If you share the set of vectors I'd be happy to implement a solver to try this.
Royi
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Hello, thanks for your comment but both the non negativity and stability criterion are to frame the problem in the context of a network. and then to capture asymmetries in the network. But positive definite assumes symmetry so I can't use this assumption. – citizenfour Jan 20 '24 at 20:06
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@citizenfour, When you say the matrix is stable, do you mean https://en.wikipedia.org/wiki/Hurwitz_matrix? Maybe a convex approximation would be negative semi definite. So $\boldsymbol{A} + \epsilon \boldsymbol{I} \in \mathcal{S}_{-}^{n}$ for a very small $\epsilon$. This means any of its eigen vector is negative. – Royi Jan 20 '24 at 21:45