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).

  • 1
    $\begingroup$ @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. $\endgroup$ Dec 31, 2017 at 13:19

1 Answer 1


There is an envelope theorem for the setting you describe.

Have a look at “Envelope Theorems for Arbitrary Choice Sets” by Milgrom and Segal (2002).


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