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.

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    $\begingroup$ 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 '16 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 '16 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 '16 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 '16 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 '16 at 22:46

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 $$ u(x,y)=K[w(x,y)]^{-1}=\frac{K}{w(x,y)}. $$


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