<p>There is an envelope theorem for the setting you describe.</p>

<p>Have a look at “<a href="http://onlinelibrary.wiley.com/doi/10.1111/1468-0262.00296/abstract" rel="noreferrer">Envelope Theorems for Arbitrary Choice Sets</a>” by Milgrom and Segal (2002).</p>

CC BY-SA 3.0
2018-01-07T06:40:37.023Z
I have tried to apply it in a very simple case. Perhaps I am confused as to the meaning of the theorem, but here is a question I asked about it: https://economics.stackexchange.com/questions/20001/does-the-envelope-theorem-hold-at-a-corner-solution
user56834
CC BY-SA 3.0
2017-12-31T12:25:38.513Z
2017-12-31T12:25:38.513Z
Theoretical Economist
<p>If we have a function
$$f(x)=\max_yg(x,y)$$</p>

<p>Then we can find the derivative $d/dx \ f(x)$ by realizing that 
$$(1): \quad \frac {\partial }{\partial y}g(x,y^*)=0$$
because of the first order condition for maximization. </p>

<p>We can use this by recognizing that $$\frac d {dx}f(x)=\frac d {dx}g(x,y^*(x))=\frac \partial {\partial x}g(x,y^*)+\frac \partial {\partial y}g(x,y^*)\frac {dy^*(x)} {dx}=\frac \partial {\partial x}g(x,y^*)$$
Where the last equality follows because of result $(1)$. </p>

<hr>

<p>However, my question is, <strong>what if $y$ can only take a discrete number of values</strong>? such as $0$ or $1$? We can of course say: </p>

<p>$$f'(x)=\begin{cases}
g_x(x,0) \quad \text { if } g(x,0) &gt; g(x,1)\\
g_x(x,1) \quad \text { if } g(x,0) &lt; g(x,1)
\end{cases}$$</p>

<p><strong>However, I am wondering if there is a kind of "envelope theorem for discrete choice sets", that would allow us to simplify this (especially if the choice set is discrete but large).</strong> </p>

CC BY-SA 3.0
2017-12-31T13:19:49.107Z
@denesp the OP wants to take the derivative with respect to the parameter $x$, and not the (discrete) choice variable $y$. Presumably a classical definition of the derivative with respect to $x$ can be applied here.
Theoretical Economist
<mathematical-economics><choice-theory>
Envelope theorem for discrete choice sets?
user56834
# Envelope theorem for discrete choice sets?

(Asked by: username_0 on 2017-12-31)

If we have a function
$$f(x)=\max\_yg(x,y)$$

Then we can find the derivative $d/dx \ f(x)$ by realizing that
$$(1): \quad \frac {\partial }{\partial y}g(x,y^\*)=0$$
because of the first order condition for maximization.

We can use this by recognizing that $$\frac d {dx}f(x)=\frac d {dx}g(x,y^\*(x))=\frac \partial {\partial x}g(x,y^\*)+\frac \partial {\partial y}g(x,y^\*)\frac {dy^\*(x)} {dx}=\frac \partial {\partial x}g(x,y^\*)$$
Where the last equality follows because of result $(1)$.

---

However, my question is, **what if $y$ can only take a discrete number of values**? such as $0$ or $1$? We can of course say:

$$f'(x)=\begin{cases}
g\_x(x,0) \quad \text { if } g(x,0) > g(x,1)\\
g\_x(x,1) \quad \text { if } g(x,0) < g(x,1)
\end{cases}$$

**However, I am wondering if there is a kind of "envelope theorem for discrete choice sets", that would allow us to simplify this (especially if the choice set is discrete but large).**

**Tags:** mathematical-economics, choice-theory
> **Comments:**
> **username_1**: @denesp the OP wants to take the derivative with respect to the parameter $x$, and not the (discrete) choice variable $y$. Presumably a classical definition of the derivative with respect to $x$ can be applied here.  

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## 1 answers

### Answer by username_1 on 2017-12-31. Score: 5 | Accepted answer

There is an envelope theorem for the setting you describe.

Have a look at “[Envelope Theorems for Arbitrary Choice Sets](http://onlinelibrary.wiley.com/doi/10.1111/1468-0262.00296/abstract)” by Milgrom and Segal (2002).

> **Comments:**
> **username_0**: I have tried to apply it in a very simple case. Perhaps I am confused as to the meaning of the theorem, but here is a question I asked about it: https://economics.stackexchange.com/questions/20001/does-the-envelope-theorem-hold-at-a-corner-solution  

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