I am reading about tranforming sharpe ratio into convex problem

After some following, its converted into min xTxy s.t. (u-rf e)x = 1


X+ = x/(u-rf e)x
K+ = 1/ (u-rf e)T x

enter image description here

In the book, it mention adding the normalizing constraint (u-rf)x = 1 does not affect the equation

I understand that f(x) is same as f(kx)

However, i get some questions here.

Why (u-rf)x = 1 can confirm its maximum sharpe ratio ?

So why this constraint

  • 4
    $\begingroup$ This is a really messy question. Formulae not written in Latex. Sentences not clearly written out. A pasted image of a large block of text. I would be surprised if someone takes this up to answer. Perhaps some editing might help... $\endgroup$
    – Attack68
    Commented Apr 29 at 15:17
  • 1
    $\begingroup$ The page 62 you reproduce is found in Optimization in Finance by Reha Tutuncu, Research Report B-392, August 2003 here citeseerx.ist.psu.edu/… but your specific question I was not able to understand $\endgroup$
    – nbbo2
    Commented Apr 29 at 16:51
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    $\begingroup$ Its mentioned adding the normalizing constraint. So my question is why to add this constraint! $\endgroup$
    – andy
    Commented Apr 30 at 4:16
  • $\begingroup$ Sorry i am typing on phone $\endgroup$
    – andy
    Commented Apr 30 at 4:19
  • 1
    $\begingroup$ Please resubmit from computer with latex formulae if possible, and clarify question. $\endgroup$
    – Sebapi
    Commented Apr 30 at 5:07

1 Answer 1


After reading about A Signal Processing Perspective on Financial Engineering.

I am trying to answer my question. But I am not sure that if I am correct or not. and show my question again.

I would like to ask why we add this (µ − rf1) = 1 constraint.

min xT Σx
subject to xT
(µ − rf1) = 1,
xT 1 > 0.

First, we do homogenizing x = kx, as f(x) = f(kx), so the objective function becomes wT Σw

x = x+ / k+
x+ = x / (µ − rf1)
k+ = 1 / (µ − rf1)

this paramater given in max, so in min k become (µ − rf1)

after x = x+ / k+, it will be same as x

as x = k+x+

as we scale constraint wT 1 = 1 can be relaxed to wT 1 > 0 so we can assume one arbitrarily set as wT(µ − rf1) = 1

k = (µ − rf1) `in min that` wT(µ − rf1) = 1 
  • $\begingroup$ Again this is confusing. By your own admission you do not know if this is an answer to your question or an addition to it. It will be difficult for anyone to follow along. $\endgroup$
    – Attack68
    Commented May 1 at 9:38

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