If I have two products $x$ and $y$ and two users $A$ and $B$ whose utility functions for those products $u$ and $w$ are inversely proportional, would it be ok to express this relationship as: $u(x,y) = w^{-1}(x,y)$? Thanks.

  • 1
    $\begingroup$ Two variables $p,q$ are inversely proportional if $pq=K$ where $K$ is a constant. So if $u$ and $w$ are inversely proportional if for any pair of products $(x,y)$, $$u(x,y)=\frac{K}{w(x,y)}$$ for some fixed $K$. $\endgroup$
    – Herr K.
    Jan 13, 2016 at 4:51
  • $\begingroup$ This is somewhat a strange relationship. It means that $A$ dislikes consuming the goods that $B$ likes. Is it what you have in mind? Otherwise, could you provide a little more context? $\endgroup$
    – Oliv
    Jan 13, 2016 at 11:19
  • $\begingroup$ @HerrK could you please explain me why you need $pq$ to define $K$? I don't know much about utility functions, but I am trying to focus on only two products $xy$. Or in your example $K$ is a constant of $xy$? $\endgroup$
    – vabm
    Jan 13, 2016 at 22:38
  • $\begingroup$ @Oliv yes, that's exactly the kind of relationship I am looking to establish. For example if I have two products that may have different colours $A$ dislikes exactly what $B$ likes and vice versa. Does $w^{-1}$ implies that? $\endgroup$
    – vabm
    Jan 13, 2016 at 22:41
  • $\begingroup$ @vabm: You can think of $p,q$ in my case as two utility numbers; for instance $u(x,y)=p$ and $w(x,y)=q$. If $p$ and $q$ are inversely proportional, then $p=K/q$, for some $K$. In your case, $K=1$, but it could take other values too. $\endgroup$
    – Herr K.
    Jan 13, 2016 at 22:46

1 Answer 1


Converting my comment to an answer...

I suppose the expression you wrote is almost correct, except perhaps for a notational issue. I would write $u(x,y)=[w(x,y)]^{-1}$, because $w^{-1}(x,y)$ is usually used to denote the pre-image of $w(x,y)$.

More generally, inverse proportionality between $u$ and $w$ is defined up to a multiplicative constant $K$; that is, $u(x,y)\cdot w(x,y)=K$ for some $K\in\mathbb R$. In the above case, you implicitly assumed $K=1$. So generally, you'd have $$ u(x,y)=K[w(x,y)]^{-1}=\frac{K}{w(x,y)}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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