The Value Function and the Envelope Trick

The Value Function

Definition: Value Function

The value function of a maximization (or minimization) problem is the value of the objective function evaluated at the solution as a function of the exogenous variables. $$V(\mathbf c,\boldsymbol\theta) =\max\limits_{\substack{\mathbf x\\ \mathrm{subject\, to} \\ \\g_1(\mathbf x)\le c_1\\[-6pt] \vdots \\[1.5pt] g_k(\mathbf x)\le c_k }} f(\mathbf x,\boldsymbol\theta)$$

The Value Function

Examples

Maximize the utility $u(x,y)=xy$ subject to the budget constraint $2x+3y\leq i$ where $i$ is the given income level. The solution is $x=i/4$ and $y=i/6$ so the value function is $V(i)=u(i/4,i/6)=i^{2}/24$. For these type of problems the value function is also called the indirect utility.
Minimize the cost of inputs $7k+2l$ subject to the quantity constraint $F(k,l)=\sqrt{kl}=Q$ where $Q$ is the given target quantity level. The solution is $k=\sqrt{2/7}Q$ and $l=\sqrt{7/2}Q$ so the value function is $V(i)=7\sqrt{2/7}Q+2\sqrt{7/2}Q=2\sqrt{14}Q$. For these type of problems the value function is also called the cost function.

You should be able to solve these problems using our recipe (and Mathematica if needed) for optimization with equality constraints. Try it and check if your solution coincides with the solution presented here.

The Envelope Trick

Consider a maximization problem with inequality constraints where the objective function depends on some exogenous variables $\mathbf c$ and $\boldsymbol\theta$. $$V(\mathbf c,\boldsymbol\theta)=\max\limits_{\substack{\mathbf x\\ \mathrm{subject\, to} \\ \\g_1(\mathbf x)\le c_1\\[-6pt] \vdots \\[1.5pt] g_k(\mathbf x)\le c_k }} f(\mathbf x,\boldsymbol\theta).$$ Let $V(\mathbf c,\boldsymbol\theta)$ be the corresponding value function of the optimization problem and let $x^\ast$ be the solution (notice that the solution also depends on the $\boldsymbol\theta$ parameter). The envelope theorem states that:

$$\nabla_{\boldsymbol\theta}V(\mathbf c,\boldsymbol\theta)= \nabla_{\boldsymbol\theta}f(\mathbf x^\ast,\boldsymbol\theta)$$

Figure 1The envelope trick
Freeze the choice at $x^\ast(\overline{\theta})$ and let $\theta$ move. The frozen objective can never beat the re-optimized value $V$, and the two meet at $\overline{\theta}$—so there they share a tangent.

Mathematica example (.nb file)

The Economic Meaning of The Lagrangian Multipliers

Shadow Prices

The value of a Lagrangian multiplier $\lambda_j^\ast$ (at the solution) is the marginal value of relaxing the constraints:

$$\diff{}{c_{j}}V(\mathbf c,\boldsymbol\theta)=\lambda_{j}^\ast$$

Put simply: the value of the objective function at the optimal will increase by $\approx \Delta c_j\times \lambda_j^\ast$ if we increase $c_j$ by $\Delta c_j$.

Proof: Consider the Lagrangian $\mathcal L(\mathbf x,\boldsymbol\lambda)$. Let $\mathbf x^\ast(\mathbf c,\boldsymbol\theta)$ and $\boldsymbol\lambda^\ast(\mathbf c,\boldsymbol\theta)$ be the solutions of the max. problem. Then $\mathcal L(\mathbf x^\ast(\mathbf c,\boldsymbol\theta),\boldsymbol\lambda^\ast(\mathbf c,\boldsymbol\theta))$ is the value function. Why?

Applying the envelope trick,

$$\diff{}{c_j}\mathcal L(\mathbf x,\boldsymbol\lambda)=\lambda_j.$$