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Trying to solve for (vertical) vector $β$ of length $n$, that maximizes scalar function $f(β)$

$$f(\beta) = \frac{\beta^T \mu}{\sqrt{\beta^T M \beta}}$$

where $μ$ is a (vertical) vector of length $n$, and $M$ is a $n \times n$ matrix.

Is there a closed form solution for $β$ - and what is it, if it exists?

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    What you call a vertical vector is more commonly referred to as a column vector.2012-11-09

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We can replace $M$ by its symmetric part $A=(M+M^\top)/2$ without changing $f$. $A$ must be positive definite for $f$ to be defined for all $\beta$. Then we can transform to $x=A^{1/2}\beta$ to obtain $f(x)=x^\top A^{-1/2}\mu/\lVert x\rVert$. Thus $f$ is maximized by any $x$ colinear with $A^{-1/2}\mu$, in particular by $x=A^{-1/2}\mu$ itself. The solution for $\beta$ in terms of $M$ is therefore

$$ \beta=A^{-1}\mu=2(M+M^\top)^{-1}\mu\;. $$

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    Thank you! - was struggling with matrix simplification2012-11-09
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    what if this optimization was constrained with $ g(\beta) = \beta^T M \beta \le K^2 $ - what would the optimal constrained solution look like?2012-11-09
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    @user1766394: If you understand the solution, you can answer that question yourself. $\beta^\top M\beta$ is just $\lVert x\rVert^2$, and as I wrote, any multiple of the solution I gave is also a solution, since the objective function is invariant under scaling, so your additional condition merely selects the multiples of the solution I gave with $\lVert x\rVert\le K$.2012-11-10
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    Sorry - an additional condition I did not mention is that each of the elements in $\beta$ has to be non-negative, in addition to the $g(\beta)$ constraint..2012-11-10
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    @user1766394: I don't bother with questions after the second unnecessary correction. Who knows what the third one will be?2012-11-10