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Do utility functions imply that if a consumer's income infinite, his consumption should also be infinite? The reason why I'd think this is the case is based on my basic understanding of utility functions.

Recall your cobb-douglas style utility function:

$$U(x_1...x_n)=\prod_{i=1}^nx_i^{\alpha_i}$$

where $0<\alpha_i<1$ and $\sum_{i=1}^n\alpha_i=1$

If our consumer is utility maximizing he should spend all of his income in a given period or even over multiple periods.

A practical application would be to answer the question of whether or not consumption growth in consumption (C) in the GDP equation can grow holding the number of consumers in an economy fixed as argued on the my answer Can an economy grow without its population growing?.

Thank you to Kenny LJ for inspiring this question.

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When the utility function is strictly increasing, it follows that more consumption provides more utility. So it seems evident to say "more income more consumption". But discussing infinity outside mathematics makes no sense. So we want to examine

$$u'(c) > 0\;\;\; \forall c,\;\;\; I\to \infty \;\;\;?\implies? \;\;\;c\to \infty$$

And the answer is "the question is indeterminate" or at least "it depends on the particulars", since what will matter is the rate at which Income and Consumption "go to infinity". If consumption goes faster, it will eventually deplete the infinite income, and if the consumer has infinite horizon, it will end up with no income. Etc.

If we want to entertain concepts from Non-Standard Analysis, and treat "infinity" not as a limiting concept but as something more concrete, then the answer will be the same, because this will be indeterminate also in Non-Standard Analysis, absent the particulars of the situation.

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  • $\begingroup$ Woulden't this be a general criticism of using static models in consumer theory? $\endgroup$ – EconJohn Jan 22 '18 at 0:43
  • $\begingroup$ @EconJohn The fact that in real life people make intertemporal choices, does not invalidate the insights that can be gained from static analysis. $\endgroup$ – Alecos Papadopoulos Jan 22 '18 at 11:06

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