Sorry if this is really easy for you gurus. I'm trying to derive the reformulation of a weighted least square regression to a quadratic programming form. I understand there is a closed form solution under some assumptions. Just to clarify, it's not a homework question, it's a reduced form of a bigger problem faced in work. Tried to google but didn't find detailed steps.

More specifically, the weighted least square regression is $min_{f} ||W(Xf-r)||^2$ and the QP form is just the standard form enter image description here

I've made some derivation as shown at bottom. I think I got $P$ but I'm not sure about $q$ due to the $f^T$ in the second term. Can you suggest if my $P$ is right and also how to get $q$?

enter image description here


1 Answer 1


Yes, your matrix $P$ is right. Note that $f^{\top} X^{\top} W^{\top} W r$ is scalar and the transpose of a scalar is just the scalar itself, i.e.

$$ f^{\top} X^{\top} W^{\top} W r = (f^{\top} X^{\top} W^{\top} W r)^{\top} = (X^{\top} W^{\top} W r)^{\top} f = r^{\top} W^{\top} W X f $$

and thus

$$ q^{\top} f = -f^{\top} X^{\top} W^{\top} W r - r^{\top} W^{\top} W X f = -2r^{\top} W^{\top} W X f. $$

So we have $q = -2(r^{\top} W^{\top} W X)^{\top} = -2X^{\top}W^{\top}W r$.

  • $\begingroup$ Ah, thank you sir! Forgot it's scalar. Thank you so much! $\endgroup$
    – inf
    Commented Jan 28, 2022 at 19:23
  • $\begingroup$ BTW, I guess the last line is $q^T$ instead of $q$. So, $q = -2 X^TW^TWr$. But got your point. Thank you so much again! $\endgroup$
    – inf
    Commented Jan 28, 2022 at 19:37
  • 1
    $\begingroup$ Yes, you're right. Good catch! $\endgroup$
    – joni
    Commented Jan 28, 2022 at 20:29

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