2
$\begingroup$

I want to maximise the function:

$$l(\beta,\sigma,\alpha) = -n\log(\sigma) - \frac{1}{\sigma} A(\alpha)\vert{\bf y}-{\bf X}\beta\vert,$$

where $\vert \cdot \vert $ represents the entry-wise absolute value, $\sigma>0$, $\alpha\in {\mathbb R}$, $\beta \in{\mathbb R}^p$, ${\bf y}\in{\mathbb R}^n$, ${\bf X}$ is an $n\times p$ real matrix, and $A(\alpha)$ is a $1\times n$ vector with positive entries which depends on the parameter $\alpha$. This kind of looks like a Linear Programming problem to me but I can't figure out how to formulate it properly as such. I can optimise this function using basically any numerical software, but it would be nice to link it to Linear Programming. I would appreciate any hints in this direction.

$\endgroup$
  • $\begingroup$ Two questions: (1) Is the way in which $A(\alpha)$ depends on $\alpha$ known? (2) By "entry-wise absolute value", might you mean the sum of the absolute values? (If not, then I don't understand what is meant?) ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 17 '15 at 18:32
  • $\begingroup$ @MichaelHardy Yes, the functional expression of $A(\alpha)$ is known. By $\vert {\bf y} -{\bf X}\beta\vert $ I meant $(\vert y_1-x_1\beta\vert, \dots, \vert y_n-x_n\beta\vert )^{\top}$. So, it is a weighted sum, given by the product of this vector and $A$. $\endgroup$ – Gillette Mar 17 '15 at 18:35
  • $\begingroup$ It is not a linear program. Far from it, as you have logarithms, products and divisions involving decision variables. $\endgroup$ – Johan Löfberg Mar 17 '15 at 19:02
  • $\begingroup$ @JohanLöfberg Many thanks for the clarification. I agree with your comment. If you post it as an answer, I will accept it. $\endgroup$ – Gillette Mar 17 '15 at 19:06
2
$\begingroup$

It is not a linear program. Far from it, as you have logarithms, products and divisions involving decision variables.

$\endgroup$
0
$\begingroup$

A couple of thoughts:

  • The values of $\alpha$ and $\beta$ that maximize this should be independent of $\sigma$. So if you find them, you can just plug them in and then find the maximizing value of $\sigma$.

  • You have $\ell = -n\log\sigma+\dfrac B \sigma$. So $$\dfrac{d\ell}{d\sigma} = \dfrac{-n}\sigma-\frac B {\sigma^2} = \dfrac{-n\sigma-B}{\sigma^2}.$$ This $=0$ if $\sigma=\dfrac{-B}n$ and $>0$ or $<0$ according as $\sigma<\text{ or }>\text{that}$. So that is the value of $\sigma$ that maximizes $\ell$. (Notice that we must have $B<0$.)

$\endgroup$
  • $\begingroup$ Very useful comments. Thanks. $\endgroup$ – Gillette Mar 19 '15 at 0:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.