I've been looking at expenditure systems and have been really interested in the behaviour of the demand system that underlies bliss points:

Consider the bliss point utility function of the following form:


for two dimensions the corresponding hicksian demands are:

$$x_1^c=\delta_1-\left[\frac{\bar{U}}{1+\frac{p_2}{p_1}}\right]^\frac{1}{2}$$ $$x_2^c=\delta_2-\left[\frac{\bar{U}}{1+\frac{p_1}{p_2}}\right]^\frac{1}{2}$$

It follows that the expenditure function is: $$e(p_1,p_2,\bar{U})=p_1\delta_1-p_1\left[\frac{\bar{U}}{1+\frac{p_2}{p_1}}\right]^\frac{1}{2}+p_2\delta_2-p_2\left[\frac{\bar{U}}{1+\frac{p_1}{p_2}}\right]^\frac{1}{2}$$

Obviously expenditure functions can be much larger. however I'm having a hard time for generating a expenditure function for a number of $n$ goods.

tldr What would the hicksian demands look like for the utility function: $$U(\mathbf{x})=-\sum_{i=1}^n(x_i-\delta_i)^2$$


2 Answers 2


Usually the change in variables $X=x-\delta$ allows to write $$ e(p,u) = \max_{x \geq 0}\{ p'x : U(x-\delta) \geq u \} = \max_{X \geq -\delta} \{ p'X : U(X) \geq u \} + p'\delta $$ and so $ e(p,u) = E(p,u) + p'\delta.$


Edit: My previous answer contained a mistake for the case where $x$ is restricted to $\mathbb{R}_{+}^{n}$. I removed this case from my answer.

Take $\bar{U} > 0$. Let's denote $\delta = (\delta_{1}, \ldots, \delta_{n})$ and $p = (p_{1}, \ldots, p_{n})$. Assume $p \neq 0$. We want to solve \begin{align*} \min_{x\in\mathbb{R}^{n}} p \cdot x \qquad \text{s.t.}\quad - (x - \delta) \cdot (x - \delta) \geq - \bar{U}. \end{align*}

Let's first note that a solution must satisfy $(x - \delta) \cdot (x - \delta) = \bar{U}$. This is because the left side of the constraint is continuous in $x$, and because every neighbourhood of $x$ in $\mathbb{R}^{n}$ contains a point $x^{\prime}$ such that $p\cdot x^{\prime} < p \cdot x$ (since $p \neq 0$).

Setting up a Lagrangian, we solve \begin{align*} \min_{x\in\mathbb{R}^{n}, \lambda\in\mathbb{R}} p \cdot x + \lambda\left( (x - \delta) \cdot (x - \delta) - \bar{U} \right). \end{align*} The first order conditions are $$ p + 2 \lambda (x - \delta) = (0, \ldots, 0) $$ and $$ (x - \delta) \cdot (x - \delta) - \bar{U} = 0. $$ Note that $\lambda$ cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to $(x - \delta) = - p / (2\lambda)$. Plugging this expression for $(x - \delta)$ into the second FOC, $$ \frac{p \cdot p}{4 \lambda^{2}} = \bar{U} $$ and hence $\lambda = \sqrt{\frac{p\cdot p}{4 \bar{U}}}$. Finally, $$ x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{\bar{U}}{p\cdot p}} $$

To gain a geometric intuition, let's denote by $\Vert\cdot\Vert$ the Eucliden norm on $\mathbb{R}^{n}$ and observe that the set of feasible choices is $$ \left\lbrace x \in\mathbb{R}^{n}\colon \Vert\delta - x \Vert \leq \sqrt{\bar{U}}\right\rbrace. $$ This is nothing but the closed ball of radiuc $\sqrt{\bar{U}}$ around $\delta$. Minimizing the linear function $x\mapsto p \cdot x$ involves finding the point on the boundary of the ball where a level set of $x\mapsto p \cdot x$ is tangent to the ball. We find this point by traveling from the ball's center at $\delta$ in the direction of $-p$, i.e. orthogonal to the hyperplane, until we reach the boundary.

Here's an alternate proof that doesn't use any stuff involving Lagrangians: As before, we may restrict attention to points $y$ satisfying $\Vert\delta - y \Vert = \sqrt{\bar{U}}$. Any such point may be written as $y = \delta - (\delta - y)\frac{\sqrt{\bar{U}}}{\Vert \delta - y \Vert}$. Consider the point $x = \delta - p \frac{\sqrt{\bar{U}}}{\Vert p \Vert}$. We claim that $p\cdot x \leq p \cdot y$ for any such point $y$ on the boundary. This is the case if and only if \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq p\cdot p \frac{1}{\Vert p \Vert}. \end{align*} The right side equals $\Vert p \Vert$. As for the left side, by the Cauchy-Schwarz inequality, \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq \Vert p \Vert \Vert \delta - y \Vert \frac{1}{\Vert \delta - y \Vert} = \Vert p\Vert, \end{align*} and so we're done. Moreover, the Cauchy-Schwarz inequality holds strictly unless $y - \delta$ is a multiple of $p$. In that case, however, either $p = \delta - y$ or $y$ does not lie on the boundary. So the point $x$ is in fact the unique minimizer.

  • $\begingroup$ Hi, looks like you're on to something. but it seems that you made a mistake.your solved FOC should be $(x-\delta)=-\frac{p}{2\lambda}$. $\endgroup$
    – EconJohn
    Nov 13, 2019 at 20:44
  • $\begingroup$ You're right, that sign is wrong, but the rest of the argument is unchanged. I'll edit the answer $\endgroup$ Nov 13, 2019 at 21:02

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