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If I do the LaGrangian for the Expenditure minimization problem, it comes as p1 = p2 = p3, how do I substitute it back in the constraint and find the Hicksian demand to find e(p,u)?

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You don’t because the utility function is linear. Trying to use the Lagrangian method on linear utility functions yields non-sensical equations. (The consumer can’t make all prices equal).

The optimization problem is

$\min p_1 x_1 + p_2 x_2 + p_3 x_3$

subject to $x_1 + x_2 + x_3 = \overline{U}$

You have to consider cases depending on which price is the least. From each case you can plug in the Hicksian demands into the expenditure $p_1 x_1 + p_2 x_2 + p_3 x_3$ to get the expenditure function $e(p_1,p_2,p3,\overline{U})$.

  • $p_1 < p_2, p_3$

$\implies x_1 = \overline{U}, x_2 = 0, x_3 = 0$

This yields $e = p_1 \overline{U}$

  • $p_2 < p_1, p_3$

$\implies x_1 = 0, x_2 = \overline{U}, x_3 = 0$

This yields $e = p_2 \overline{U}$

  • $p_3 < p_1, p_2$

$\implies x_1 = 0, x_2 = 0, x_3 = \overline{U}$

This yields $e = p_3 \overline{U}$

  • $p_1 = p_2 < p_3$

$\implies x_1 = \alpha, x_2 = \overline{U} - \alpha, x_3 = 0, 0 \leq \alpha \leq \overline{U}$ (a line in 3D space)

This yields $e = p_1 \overline{U} = p_2 \overline{U}$

  • $p_1 = p_3 < p_2$

$\implies x_1 = \alpha, x_2 = 0, x_3 = \overline{U} - \alpha, 0 \leq \alpha \leq \overline{U}$ (a line in 3D space)

This yields $e = p_2 \overline{U} = p_3 \overline{U}$

  • $p_2 = p_3 < p_1$

$\implies x_1 = 0, x_2 = \alpha, x_3 = \overline{U} - \alpha, 0 \leq \alpha \leq \overline{U}$ (a line in 3D space)

This yields $e = p_2 \overline{U} = p_3 \overline{U}$

  • $p_1 = p_2 = p_3$

$\implies x_1 = \alpha, x_2 = \beta, x_3 = \overline{U} - \alpha - \beta, 0 \leq \alpha \leq \overline{U}, 0 \leq \beta \leq \overline{U} - \alpha$ (a plane in 3D space)

This yields $e = p_1 \overline{U} = p_2 \overline{U} = p_3 \overline{U}$

Edit: In general for this problem, the expenditure function looks like this:

$e(p_1,p_2,p_3,\overline{U}) = p_{min} \overline{U}$, where

$p_{min} = \min\{p_1,p_2,p_3\}$.

Note the optimal expenditure function we got $e(p_1,p_2,p_3,\overline{U})$ is piecewise defined, hence not differentiable. This is the reason why you can’t get it through the Lagrangian method, which relies on derivatives (hence differentiability).

The above always happens when given a linear utility function.

Note: $\#$ parameters $=$ dimension of the optimal bundles manifold

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