Demand Functions Derivations

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### Hicksian (Compensated) Demand Derivation Hicksian demand functions minimize expenditure for a fixed utility level. #### 1. Expenditure Minimization Problem - **Objective Function (Expenditure):** $E = P_xX + P_yY$ - **Constraint (Utility):** $U = X^{1/2} Y^{1/2}$ #### 2. Lagrangian Equation $$\mathcal{L} = P_xX + P_yY - \lambda(U - X^{1/2} Y^{1/2})$$ #### 3. First-Order Derivatives - **With respect to X:** $P_x - \frac{1}{2}\lambda X^{-1/2}Y^{1/2} = 0 \Rightarrow P_x = \frac{1}{2}\lambda X^{-1/2}Y^{1/2}$ - **With respect to Y:** $P_y - \frac{1}{2}\lambda X^{1/2}Y^{-1/2} = 0 \Rightarrow P_y = \frac{1}{2}\lambda X^{1/2}Y^{-1/2}$ - **With respect to $\lambda$:** $U - X^{1/2}Y^{1/2} = 0 \Rightarrow U = X^{1/2}Y^{1/2}$ #### 4. Solve for the Price Ratio Divide the $P_x$ equation by the $P_y$ equation: $$\frac{P_x}{P_y} = \frac{\frac{1}{2}\lambda X^{-1/2}Y^{1/2}}{\frac{1}{2}\lambda X^{1/2}Y^{-1/2}} = \frac{Y^{1/2}Y^{1/2}}{X^{1/2}X^{1/2}} = \frac{Y}{X}$$ - Cross-multiply: $P_xX = P_yY$ - Isolate X and Y: - $X = \frac{P_yY}{P_x}$ - $Y = \frac{P_xX}{P_y}$ #### 5. Substitute Back into Utility Constraint - **Solving for $X_h$:** Substitute $Y = \frac{P_xX}{P_y}$ into $U = X^{1/2} Y^{1/2}$: $U = X^{1/2} (\frac{P_xX}{P_y})^{1/2} = X^{1/2} X^{1/2} (\frac{P_x}{P_y})^{1/2} = X (\frac{P_x}{P_y})^{1/2}$ Multiply by the reciprocal: $$X_h = U (\frac{P_y}{P_x})^{1/2}$$ - **Solving for $Y_h$:** Substitute $X = \frac{P_yY}{P_x}$ into $U = X^{1/2} Y^{1/2}$: $U = (\frac{P_yY}{P_x})^{1/2} Y^{1/2} = Y^{1/2} Y^{1/2} (\frac{P_y}{P_x})^{1/2} = Y (\frac{P_y}{P_x})^{1/2}$ Multiply by the reciprocal: $$Y_h = U (\frac{P_x}{P_y})^{1/2}$$ ### Marshallian (Uncompensated) Demand Derivation Marshallian demand functions maximize utility subject to a budget constraint. #### 1. Utility Maximization Problem - **Objective Function (Utility):** $U = X^{1/2} Y^{1/2}$ - **Constraint (Budget):** $M = P_xX + P_yY$ #### 2. Lagrangian Equation $$\mathcal{L} = X^{1/2} Y^{1/2} - \lambda(P_xX + P_yY - M)$$ #### 3. First-Order Derivatives - **With respect to X:** $\frac{1}{2}X^{-1/2}Y^{1/2} - \lambda P_x = 0 \Rightarrow P_x = \frac{X^{-1/2}Y^{1/2}}{2\lambda}$ - **With respect to Y:** $\frac{1}{2}X^{1/2}Y^{-1/2} - \lambda P_y = 0 \Rightarrow P_y = \frac{X^{1/2}Y^{-1/2}}{2\lambda}$ - **With respect to $\lambda$:** $M - P_xX - P_yY = 0 \Rightarrow M = P_xX + P_yY$ #### 4. Solve for the Relationship Between X and Y Divide the $P_x$ equation by the $P_y$ equation: $$\frac{P_x}{P_y} = \frac{\frac{X^{-1/2}Y^{1/2}}{2\lambda}}{\frac{X^{1/2}Y^{-1/2}}{2\lambda}} = \frac{X^{-1/2}Y^{1/2}}{X^{1/2}Y^{-1/2}} = \frac{Y^{1/2}Y^{1/2}}{X^{1/2}X^{1/2}} = \frac{Y}{X}$$ - Cross-multiply: $P_xX = P_yY$ - Isolate X and Y: - $X = \frac{P_yY}{P_x}$ - $Y = \frac{P_xX}{P_y}$ #### 5. Substitute Back into Budget Constraint - **Solving for $X_m$:** Substitute $Y = \frac{P_xX}{P_y}$ into $M = P_xX + P_yY$: $M = P_xX + P_y(\frac{P_xX}{P_y})$ $M = P_xX + P_xX$ $M = 2P_xX$ $$X_m = \frac{M}{2P_x}$$ - **Solving for $Y_m$:** Substitute $X = \frac{P_yY}{P_x}$ into $M = P_xX + P_yY$: $M = P_x(\frac{P_yY}{P_x}) + P_yY$ $M = P_yY + P_yY$ $M = 2P_yY$ $$Y_m = \frac{M}{2P_y}$$