Consider the standard second order cone programming problem:


\begin{equation}
\begin{array}{ll}
\operatorname{maximize} & \bar{p}^{T} x \\
\text { subject to } & \bar{p}^{T} x+\Phi^{-1}(\beta)\left\|\Sigma^{1 / 2} x\right\|\_2 \geq \alpha.
\end{array}
\end{equation}


This is a problem appearing in portfolio optimization ([here](https://r-forge.r-project.org/scm/viewvc.php/*checkout*/pkg/Rsocp/inst/doc/socp.pdf?root=rmetrics) for more info).
Here we assume $\bar{p},x \in\mathbb{R}^n$ and $\alpha, \beta\in[0,1]$, $\Sigma \in \mathbb{R}^{n\times n}$ is Positive Semi-Definite Matrix and $\Phi^{-1}(\cdot)$ is the inverse Gaussian CDF. The problem is convex for $\beta\leq0.5$.


Is there any closed-form solution for this problem? I tried solving the problem by imposing KKT but it did not lead to anything promising. Thanks.



Convex optimization closed-form solution
For this special case an explicit solution is possible. To simplify notation let $A := \Sigma^{1/2}$ and let $A^+$ be the pseudoinverse of $A$ (see f.i. Golub/Van Loan: Matrix Computations (1996), 3rd ed., p. 257). Let $y := Ax$, $z := (E - A^+A)x$, thus $x = A^+y +z$ and the problem is
$$p^T(A^+y + z) = max!\\p^T(A^+y+z) + \Phi^{-1}(\beta) \cdot \sqrt{y^TAA^+y} \geq \alpha,$$
since $Az = 0$ and $(AA^+y)^T(AA^+y) = y^TAA^+y$. Now assume that $\beta < 0.5$ (the case $\beta = 0.5$ is uninteresting since then $\Phi^{-1}(\beta) = 0$). Let $F$ be the image of $x \to Ax$ and $\mathbb{R}^n = F+G$, $G$ a subspace. If there is some $z \in G$ with $p^T z \not= 0$, then the problem is unsolvable (take $y = 0$ and $x = \lambda z$ with either $\lambda < 0$ or $\lambda > 0$). Of course this only can happen if $A$ is singular.


Now we can assume w.l.o.g. that $A$ is non-singular. Then $AA^+ = E$, the identity matrix and always $z = 0$. Let $q := (A^+)^Tp$, then the problem is
$$q^T y = max!\\q^T y + \Phi^{-1}(\beta) \cdot \sqrt{y^Ty} \geq \alpha$$
This problem is much easier to solve. Simply let $y = \lambda q$. Then the problem is
$$\lambda q^Tq = max!\\ \lambda q^Tq + \Phi^{-1}(\beta) \cdot |\lambda| \cdot \|q\|\_2 \geq \alpha$$
Which can be solved directly. Also general cases (singular $A$), which are solvable, can be reduced to this special case.


