# CES utility with negative values

We know that the CES utility function approximates a Leontief utility function as in the following:

$$u(x_1,x_2) = (x_1^{-r}+x_2^{-r})^{-1/r}\overset{r\rightarrow\infty}{\longrightarrow}\min\{x_1, x_2\}$$

However, this limit only holds for $$x_i\ge0$$. Is there a CES-like utility function, which also holds for negative $$x_i$$ and large $$r$$?

I am particularly interested in functions of the form $$u(f_1(x_1),f_2(x_2))$$ with $$f_i(x_i) = \frac{x_i^{1-\eta_i}-1}{1-\eta_i}, \eta_i>0$$. Adding a large constant to $$f_i$$ is not feasible since $$f_i$$ can get arbitrarily small (at least for $$\eta_i\ge1$$). So for $$f_i(x_i)<0$$, the function $$u(f_1(x_1),f_2(x_2))$$ does not behave like $$x_1$$ and $$x_2$$ are complements. Is there a different formulation of CES utility which does not have this problem?

• You want to be able to hold negative amounts of $x_i$, but do you also want to have positive amounts? If everyone wants negative amounts, if $x_i$ is a "bad", maybe you can just define $y_i=-x_i$ and proceed as normal.
– BKay
Aug 13 '21 at 13:55
• Yes, unfortunately positive and negative values have to be accounted for. In my case, I am constructing a nested utility function, so the interpretation of $x_i$ as good or "bad" is not so straightforward anymore. If this family of utility functions would be defined only on one half plane, some trickery might be possible. But here, $f_i(x_i)$ are defined on all $\mathbb{R}$... Aug 16 '21 at 11:13
• Elasticities are hard to interpret in situations where quantities are sometimes negative and sometimes positive. See for example this question: economics.stackexchange.com/questions/29696/…
– BKay
Aug 16 '21 at 14:08
• That's good to know, thanks! But does it prevent us from defining sensible elasticities of substitution for these cases? Or is it inherently impossible because we have $$u(x_1,x_2) \overset{\eta\rightarrow0^+}\longrightarrow\min\{x_1,x_2\},\\ u(x_1,x_2) \overset{\eta\rightarrow0^-}\longrightarrow\max\{x_1,x_2\}$$ (for $-r = \frac{\eta-1}{\eta}$)? Aug 19 '21 at 11:12