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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 $u(x^{\ast})>u^{\ast}$?

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

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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 $x' \in h(p,u^{\ast})$

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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 $p\cdot\big(\alpha 0+(1-\alpha)x\big)=p\cdot(1-\alpha)x=(1-\alpha) p\cdot x<p\cdot x$. So a bundle that gives higher utility than $u^*$ cannot be expenditure minimizing, and every expenditure minimizing bundle must give utility exactly $u^*$.

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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 $u(0,0)=1>0.5=u^*$.

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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.

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