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The utility function for perfect substitutes is defined as U(X,Y) = aX + bY. If the two goods X&Y are imperfect substitutes what would be their utility function?

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In general, any utility function that produces non-linear and downward sloping indifference curves will feature imperfect substitutability between goods. (By considering only goods, we're implicitly assuming "more is preferred to less", or positive marginal utility, for each good.) In other words, look at the marginal rate of substitution and see if it depends on the quantity of either good; if it does, then there is imperfect substitutability.

Examples

  • Quasi-linear utility: $u(x,y)=v(x)+y$, where $v(x)$ is non-linear in $x$ (e.g. $v(x)=\sqrt x$)
  • Additively separable utility: $u(x,y)=g(x)+h(y)$, where $g,h$ are non-linear functions
  • Cobb-Douglas: $u(x,y)=x^\alpha y^\beta$, where $\alpha,\beta$ are constants
  • Leontief (perfect complement): $\style{text-decoration:line-through}{u(x,y)=\min\{\alpha x,\beta y\}}$ [Leontief utility function features nonsubstitutability rather than imperfect substitutability.]
  • CES utility: $u(x,y)=\left(\alpha x^\rho+\beta y^\rho\right)^{1/\rho}$, where $\rho\ne1$. Note that the class of CES utility functions includes Cobb-Douglas (as $\rho\to0$), Leontief (as $\rho\to-\infty$), and linear utility (as $\rho=1$) as special cases.

This lecture note by Simon Board has examples and indifference curves corresponding to the cases above.

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  • $\begingroup$ Leontief utilities represent imperfect substitution? Not non-substitution? $\endgroup$ – Giskard Feb 20 at 6:20
  • $\begingroup$ Similary, I don't think that if $U(x,y) = x -y $ then goods $x$ and $y$ are substitutes. Having a negative slope for the indifference curve seems like a logical think to want. $\endgroup$ – Giskard Feb 20 at 6:21
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    $\begingroup$ @denesp: Thanks. Your point on Leontief utility is well taken. I've also added an explanation making clear that we're considering only goods here, which would rule out the case of negative marginal utility. $\endgroup$ – Herr K. Feb 20 at 16:57

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