Regarding the Expenditure Function Underlying a Bliss Point

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:

$$U(x_1,x_2)=-(x_1-\delta_1)^2-(x_2-\delta_2)^2$$

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

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.

• Hi, looks like you're on to something. but it seems that you made a mistake.your solved FOC should be $(x-\delta)=-\frac{p}{2\lambda}$.
– EconJohn
Nov 13 '19 at 20:44
• You're right, that sign is wrong, but the rest of the argument is unchanged. I'll edit the answer Nov 13 '19 at 21:02