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Topic: expenditure function
Conf: Chapter A2, Calculus and Optimization, Msg: 14988
From: Martin Caley (martin.caley@economics.treasury.gov.im)
Date: 10/6/2004 10:32 AM
expenditure function Martin Caley MCaley martin.caley@economics.treasury.gov.im
Another proof! This uses properties of lower bounds.
By definition e(p,u)<=p.x for all U(x)>=u. So for t>0, te(p,u)<=tp.x for all U(x)>=u. Hence te(p,u)<=e(tp,u).
Also e(tp,u)<=tp.x=t(p.x)for all U(x)>=u and hence e(tp,u)<=te(p,u). So te(p,u)<=e(tp,u)<=te(p,u) and hence e(tp,u)=te(p,u).