
## The Envelope Condition

##### You've got this Bellman's Equation Right Here:

$V(k) = \max_{k^\prime \in \Gamma(k)} \left[ \ln (Ak^\alpha-k^\prime) +\beta V(k^\prime) \right]$

You want to characterize , so you take the first order condition and the envelope condition. The First Order condition is pretty straightforward; you know this stuff: $0 = {\partial \over \partial k^\prime} \left[ \ln(Ak^\alpha-k^\prime) +\beta V(k^\prime) \right] = {-1 \over Ak^\alpha-k^\prime} + \beta V^\prime (k^\prime ) \mytag{FOC}$ And the envelope condition is: $V^\prime (k) = { A \alpha k^{\alpha-1} \over Ak^\alpha-k^\prime} \mytag{Envelope}$ Combine these two conditions to get: ${1 \over Ak^\alpha-k^\prime} = \boxed{\beta V^\prime (k^\prime ) = {V^\prime (k) \over A\alpha k^{\alpha-1}} }$ which can be used to characterizew the policy function for capital. But wait, back up, what's this about envelopes?

##### A quick math refresher

Recall from way back when you learned calculus that the multivarible version of the chain rule goes something like this: If $$f:\mathbb{R^2}\to\mathbb{R}$$ and $$(g_1,g_2)\equiv g: \mathbb{R}\to \mathbb{R}^2$$ are differentiable functions at $$x$$, and $$v(x)\equiv f(g_1(x),g_2(x)$$ then taking the total derivative, we get: ${\partial \over \partial x} v(x) = {\partial \over \partial x}g_1(x) \cdot {\partial \over \partial g_1} f(g_1,g_2) + {\partial \over \partial x}g_2(x) \cdot {\partial \over \partial g_2} f(g_1,g_2)$

In particular, let $$g_1(x) = x$$. Then $${\partial x \over \partial x }=1$$. So ${\partial \over \partial x} v(x) = {\partial \over \partial x} f(x,g_2) + {\partial \over \partial x}g_2(x) \cdot {\partial \over \partial g_2} f(x,g_2) \mytag{*}$

##### Now back to optimization

Suppose you have a problem of the form: $v(x) = \max_{y\in Y} \left[ F(x,y) + \beta V(y) \right]$ Let's take the derivative with respect to x of both sides of this equation. But how do we deal with the right-hand-side maximization doodad?

Assume that for any $$x$$, there exists at least $$y\in Y$$ such that $$y$$ solves the maximization problem given $$x$$. Let $$\bar{y}(x)$$ be a policy function mapping values of $$x$$ to these optimal $$y$$s.

Then we can reexpress the maximizing whatzit as $v(x) = F(x,\bar{y}(x)) + \beta V(\bar{y}(x))$ And so applying (*), (with $$g(x) \equiv (x,\bar{y}(x))$$ and $$f(g_1,g_2) \equiv F(g_1,g_2) + \beta V(g_2)$$): ${\partial \over \partial x} v(x) = {\partial \over \partial x} \left[ F(x,\bar{y}) + \beta V(\bar{y}) \right] + \color {RedViolet} {\partial \over \partial x} \bar{y}(x) \cdot{\partial \over \partial \bar{y}} \left[ F(x,\bar{y}) + \beta V(\bar{y}) \right]$

We've almost made it to something resembling the envelope condition! But there's a big $${\color{RedViolet} \text{mess}}$$ hanging out there on the end, getting in the way. Remember, for every $$x$$, $$\bar{y}(x)$$ solves $$\max_{y\in Y} \left[ F(x,y) + \beta V(y) \right]$$. Assuming an interior solution, this means that if $$\bar{y}=\bar{y}(x)$$, then $0 = {\partial \over \partial \bar{y}} \left[ F(x,\bar{y}) + \beta V(\bar{y}) \right] \mytag{FOC}$ And look at that. In the optimum, that mess goes away.

So the Envelope Condition says that at the optimum, the marginal effect of changing $$x$$ manifests only through the direct consequences of changing $$x$$, and not indirectly from the resulting change in $$y$$: $\boxed{ {\partial \over \partial x} v(x) = {\partial \over \partial x} \left[ F(x,\bar{y}) + \beta V(\bar{y}) \right] } \mytag{Envelope}$

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