# Expenditure min problem

The typical expenditure min. problem wants to minimize expenditure under the constraint $$u(x) \ge u^{\ast}$$. Why the solution of this problem is such that $$u(x^{\ast})=u^{\ast}$$ and not $$u(x^{\ast})>u^{\ast}$$?

• If $u(x^*)> u^*$ you might think you could reduce consumption and expenditure marginally to some $x^\prime$ and still have $u(x^\prime) \ge u^*$ Feb 7, 2023 at 10:32

Under the assumption that the utility function is continuous and represents preference relations defined on the consumption set $$X=R^{L}_+$$, and for $$p >>0$$, the hicksian demand correspondence, $$h(p,u)$$, possesses the property of non-excess utility.

$$h(p,u)$$ is the set of optimal consumption bundles solving the expenditure minimization problem (EMP) at a price vector $$p$$ and a fixed level of utility $$u$$.

I'm going to prove why a solution of the EMP, i.e., $$x \in h(p,u)$$, is such that $$u(x)=u$$, and $$x$$ is simply a consumption bundle solving the EMP.

Suppose there exists an $$x' \in h(p,u^{\ast})$$ such that $$u(x')>u^{\ast}$$, so $$x'$$ is a solution of the EMP but delivers a level of utility larger than $$u^{\ast}$$. Take a scale-down version of $$x'$$, say $$x''=\alpha x'$$, where $$\alpha \in (0,1)$$. For $$\alpha$$ sufficiently close to $$1$$, by continuity of preferences, it must be that $$u(x'') \ge u^{\ast}$$, but now $$p \cdot x'' < p x'$$, and this contraddicts $$x'$$ of being optimal in the EMP, and this is not possible. That's why $$u(x')=u^{\ast}$$ if $$x' \in h(p,u^{\ast})$$

In general, it is well possible that $$u(x^*)>u^*$$ for an expenditure minimizing $$x^*$$. Typical assumptions that guarantee $$u(x^*)=x^*$$ are that $$u$$ has domain $$\mathbb{R}^l_+$$, is continuous, $$p\gg 0$$, and $$u^*>0$$.

Here is the argument: Let $$x$$ satisfy $$u(x)>u^*>0$$. We must have $$x\neq 0$$ and $$p\cdot x>0$$. Continuity of $$u$$ implies that for $$\alpha\in(0,1)$$ small enough, $$u\big(\alpha 0+(1-\alpha)x\big)>u^*$$. Since $$p\gg 0$$, we also have $$p\cdot\big(\alpha 0+(1-\alpha)x\big)=p\cdot(1-\alpha)x=(1-\alpha) p\cdot x. So a bundle that gives higher utility than $$u^*$$ cannot be expenditure minimizing, and every expenditure minimizing bundle must give utility exactly $$u^*$$.

Consider the utility function $$u:\mathbb{R}^2_+\rightarrow\mathbb{R}$$, given by $$u(x, y) = \min(x, y) + 1$$. Consider $$u^*=0.5$$, solution to the expenditure minimisation problem when $$p_X>0, p_Y>0$$ will be $$(0,0)$$ and it satisfy $$u(0,0)=1>0.5=u^*$$.

Another example is with discontinuous utility $$u:\mathbb{R}^2_+\rightarrow\mathbb{R}$$, given by $$u(x, y) = \lfloor x+y\rfloor$$ i.e. greatest integer less than or equal to $$x+y$$. Consider $$u^*=4.2$$, $$p_X = 1$$, $$p_Y = 2$$. Solution to the expenditure minimisation problem is $$(5,0)$$ and it satisfy $$u(5,0) = 5>4.2=u^*$$.

Another example is when the two commodities are bad. Consider utility $$u:\mathbb{R}^2_+\rightarrow\mathbb{R}$$, given by $$u(x, y) = 1-x-y$$. Consider $$u^*= 0.5$$, solution to the expenditure minimisation problem when $$p_X>0, p_Y>0$$ will be $$(0,0)$$ and it satisfy $$u(0,0)=1>0.5=u^*$$.

If the preferences are continuous and monotone, if you want to get a higher utility level $$U’> U^\star$$, you need to spend more money than the optimal expenditure for the current utility level $$U^\star$$.

In functional notation, $$e(p_x,p_y,U’) > e(p_x,p_y,U^\star)$$, so you couldn’t attain the minimum for $$U’ > U^\star$$.

Taking this into account helps simplify the problem because optimizing for $$u(x^\star) = u^\star$$ can be done with a simple Lagrangian, while for $$u(x^\star) \geq u^\star$$ you need to work with the Kuhn-Tucker conditions which are a bit more complex.