Assume $U(x,y,a,c )= - c x + B(x,y,a)$, with $\frac{\partial B(x,y,a)}{\partial c }=0$, and with $a$ and $c\geq 0$ being parameters, and with $x$ and $y$ being variables. Further, $B(x,y,a)$ is continuous and twice continuously differentiable in all its inputs. Further, the $(x,y)$ maximizing $U(x,y,a,c )$ is unique everywhere and it is continuous in $a$ and $c$.

Suppose that $(x_1,y_1)$ maximizes $U(x,y,a_1,c )$, and $(x_2,y_2)$ maximizes $U(x,y,a_2,c )$, and that $0\leq x_1<x_2$.

Let $U^*(a):=\underset{x,y}{\max} U(x,y,a,c)$.

To show: $\frac{\partial U^*(a_1)}{\partial c } > \frac{\partial U^*(a_2)}{\partial c } $


1 Answer 1


It seems that you can just define two objective problems with the features you stated above and just take the derivatives with respect to $c$ to illustrate this.

Under your specification we have two value functions:

$$U(x_1,y_1,a_1,c)=-cx_1+B(x_1,y_1,a_1)$$ $$U(x_2,y_2,a_2,c)=-cx_2+B(x_2,y_2,a_2)$$

Taking the derivative of the above two equations with respect to $c$ we get:

$$\frac{\partial U(x_1,y_1,a_1,c)}{\partial c}=-x_1$$ $$\frac{\partial U(x_2,y_2,a_2,c)}{\partial c}=-x_2$$

Based on the assumption that $0\leq x_1<x_2$ it follows that $\frac{\partial U(x_1,y_1,a_1,c)}{\partial c}>\frac{\partial U(x_2,y_2,a_2,c)}{\partial c}$.

  • 1
    $\begingroup$ +1 but maybe you can add that you use the envelope theorem to obtain the expression of the partial derivatives. $\endgroup$
    – tdm
    Oct 13 at 7:47
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    $\begingroup$ @EconJohn: Thanks for your answer. The problem is that $x_1$ and $x_2$ are also functions of $c$. Why can we ignore them when taking the derivative? $\endgroup$
    – Paul
    Oct 13 at 9:32
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    $\begingroup$ Combing it with @tdm: The envelope theorem says that we can "ignore" $x^*$'s and $y^*$ change in $c$. $\endgroup$
    – Paul
    Oct 13 at 9:45
  • $\begingroup$ @Paul, Correct there's some envelope theorem applied here implicitly. $\endgroup$
    – EconJohn
    Oct 15 at 0:10
  • 1
    $\begingroup$ Thanks to both of you! $\endgroup$
    – Paul
    Oct 16 at 11:26

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