Misologia Wiki
Edit Page
Warning: You are not logged in. Your IP address will be publicly visible if you make any edits. If you log in or create an account, your edits will be attributed to your username, along with other benefits.

The edit can be undone. Please check the comparison below to verify that this is what you want to do, and then publish the changes below to finish undoing the edit.

Latest revision Your text
Line 1: Line 1:
   
In Consumer Theory, the [[Hicksian demand]] function can be related to the expenditure function by
 
   
<math>h_{i}({\mathbf{p}}, u) = \partial e(\mathbf{p}, u)/\partial p_{i} \,</math>
 
   
  +
Analogously, in Producer Theory, the [[Conditional factor demand]] function can be related to the cost function by
 
  +
<math>x_{i}({\mathbf{w}},y) = \partial c(\mathbf{p}, y)/\partial p_{i} \,</math>
 
  +
  +
  +
 
5) The Hicksian demand function can be related to the expenditure function by
  +
 
<math>h_{i}({\mathbf{p}}, u) = \partial e(\mathbf{p}, u)/\partial p_{i} \,</math>
   
 
== Derivation ==
 
== Derivation ==
The following derivation is for relationship between the Hicksian demand and the expenditure function. The derivation for conditional factor demand and the cost function is identical, only with other notation.
 
   
 
1) The expenditure function returns the cost-minimizing value for reaching a fixed level of utility <math>u \,</math> at given prices <math>\mathbf{p}</math>. As such, it is the result of a constrained optimization problem:
 
1) The expenditure function returns the cost-minimizing value for reaching a fixed level of utility <math>u \,</math> at given prices <math>\mathbf{p}</math>. As such, it is the result of a constrained optimization problem:
   
<math>\mathcal{L}(\mathbf{x}) = px-\lambda (U(x)-u)</math> with the FOCs:
+
<math>\mathcal{L}(\mathbf{x}) = px-\lambda (U(x)-u)</math> with the FOCs:
   
<math>\partial \mathcal{L}/\partial{x_{i}} = 0 \leftrightarrow p_{i}=\lambda \partial U(x)/ \partial x_{i} </math> for <math> i=1,...k \,</math>
+
<math>\partial \mathcal{L}/\partial{x_{i}} = 0 \leftrightarrow p_{i}=\lambda \partial U(x)/ \partial x_{i} </math> for <math> i=1,...k \,</math>
   
<math>\partial \mathcal{L}/ \partial \lambda = 0 \leftrightarrow U(x) = u</math>
+
<math>\partial \mathcal{L}/ \partial \lambda = 0 \leftrightarrow U(x) = u</math>
   
 
where the resulting optimal <math>\mathbf{x\,}</math> are given by <math>x_{i}^{*}=h_{i}(p_{i},u) \,</math> and the cost-minimizing expenditure is given by <math>e(p,U)=h(p,u) \cdot p \,</math>
 
where the resulting optimal <math>\mathbf{x\,}</math> are given by <math>x_{i}^{*}=h_{i}(p_{i},u) \,</math> and the cost-minimizing expenditure is given by <math>e(p,U)=h(p,u) \cdot p \,</math>
   
2) Differentiate the optimal expenditure function with respect to <math>p_{i} \, </math> using the Chain rule:
+
2) Differentiate the optimal expenditure function with respect to <math>p_{i} \,</math> using the Chain rule:
   
<math>\partial e(p,u) / \partial p_{i} = \partial h_{i}(\mathbf{p},u)\cdot p_{i} + h_{i}(\mathbf{p}, u)</math>
+
<math>\partial e(p,u) / \partial p_{i} = \partial h_{i}(\mathbf{p},u)\cdot p_{i} + h_{i}(\mathbf{p}, u)</math>
   
 
3) Replace <math>p_{i}\,</math> by the FOC of the original optimization problem:
 
3) Replace <math>p_{i}\,</math> by the FOC of the original optimization problem:
   
<math>\partial e(\mathbf{p},u) / \partial p_{i} = \partial h_{i}(\mathbf{p},u) \cdot \lambda \partial U(x)/ \partial x_{i} + h_{i}(\mathbf{p},u)</math>
+
<math>\partial e(\mathbf{p},u) / \partial p_{i} = \partial h_{i}(\mathbf{p},u) \cdot \lambda \partial U(x)/ \partial x_{i} + h_{i}(\mathbf{p},u)</math>
   
 
4) Differentiate the budget constraint FOC from (1) with respect to <math>x_{i}\,</math>:
 
4) Differentiate the budget constraint FOC from (1) with respect to <math>x_{i}\,</math>:
   
<math> \partial U(x)/\partial x_{i} = 0\,</math>
+
<math> \partial U(x)/\partial x_{i} = 0\,</math>
   
 
5) Inserting this optimality condition into (3), we get the result:
 
5) Inserting this optimality condition into (3), we get the result:
   
<math> h_{i}(\mathbf{p},u) =\partial e(p,u)/ \partial p_{i} </math>
+
<math> h_{i}(\mathbf{p},u) =\partial e(p,u)/ \partial p_{i} </math>
[[Category:Microeconomics]]
 
Please note that all contributions to the Misologia Wiki are considered to be released under the CC-BY-SA
Cancel Editing help (opens in new window)