<p>As Jesper Hypel has pointed out in a comment, the derivative need not exist. Actually, the function <span class="math-container">$a\mapsto x^*(a)$</span> need not even be nondecreasing. Here is an example: Let <span class="math-container">$g$</span> and <span class="math-container">$h$</span> be given by <span class="math-container">$g(x,a)=-(a-x)^2$</span> and <span class="math-container">$h(x,a)=a-(a-1-x)^2$</span>, respectively. Then <span class="math-container">$x_1(a)=a$</span> and <span class="math-container">$x_2(a)=a-1$</span>. Moreover, <span class="math-container">$f\big(x^*(a),a\big)=g\big(x_1(a),a\big)$</span> for <span class="math-container">$a&lt;0$</span> and <span class="math-container">$f\big(x^*(a),a\big)=h\big(x_2(a),a\big)$</span> for <span class="math-container">$a&gt;0$</span>. As a consequence, the function <span class="math-container">$a\mapsto x^*(a)$</span> jumps down after <span class="math-container">$0$</span>.</p>

CC BY-SA 4.0
2021-03-21T02:57:12.677Z
Thanks. Would there be a more definite result if h(x,a) = h(x)?
Adam
CC BY-SA 4.0
2021-03-21T08:02:33.290Z
Not really: Let $h(x,a)=-x^2$ and $g(x,a)=a-(a-1-x)^2$.
Michael Greinecker
CC BY-SA 4.0
2021-03-21T01:18:53.723Z
2021-03-21T18:25:59.883Z
2021-03-21T18:25:59.883Z
Michael Greinecker
<p>Suppose that <span class="math-container">$f(x,a)=\max[g(x,a),h(x,a)]$</span> and <span class="math-container">$x^{∗}=\arg\max_{x}f(x,a)$</span>.
Also, let <span class="math-container">$x_1(a)=\arg\max_{x}g(x,a)$</span> and <span class="math-container">$x_2(a)=\arg\max_{x}h(x,a)$</span>.</p>
<p>If both <span class="math-container">$x'_1(a)$</span> and <span class="math-container">$x'_2(a)$</span> are nonnegative, under what conditions is <span class="math-container">$\frac{dx^∗}{da}\ge 0$</span>?</p>

CC BY-SA 4.0
2021-03-21T00:39:58.370Z
I think this is better asked on mathematics. Intuitively I would guess that the derivative must be nonnegative if it exists. Question is where it exists. Places where it may not exist are places where the two functions g(x,a) and h(x,a) change from one being larger than the other. So one condition would be that g(x,a) is everywhere larger than h(x,a). But that condition is probably too strong for you. What conditions you want to impose may depend on the economic problem you are trying to model. So maybe only you can answer the question...
Jesper Hybel
<mathematical-economics>
Comparative statics of a maximum
Adam
# Comparative statics of a maximum

(Asked by: username_0 on 2021-03-20)

Suppose that $f(x,a)=\max[g(x,a),h(x,a)]$ and $x^{∗}=\arg\max\_{x}f(x,a)$.
Also, let $x\_1(a)=\arg\max\_{x}g(x,a)$ and $x\_2(a)=\arg\max\_{x}h(x,a)$.

If both $x'\_1(a)$ and $x'\_2(a)$ are nonnegative, under what conditions is $\frac{dx^∗}{da}\ge 0$?

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: I think this is better asked on mathematics. Intuitively I would guess that the derivative must be nonnegative if it exists. Question is where it exists. Places where it may not exist are places where the two functions g(x,a) and h(x,a) change from one being larger than the other. So one condition would be that g(x,a) is everywhere larger than h(x,a). But that condition is probably too strong for you. What conditions you want to impose may depend on the economic problem you are trying to model. So maybe only you can answer the question...  

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## 1 answers

### Answer by username_2 on 2021-03-21. Score: 7

As Jesper Hypel has pointed out in a comment, the derivative need not exist. Actually, the function $a\mapsto x^\*(a)$ need not even be nondecreasing. Here is an example: Let $g$ and $h$ be given by $g(x,a)=-(a-x)^2$ and $h(x,a)=a-(a-1-x)^2$, respectively. Then $x\_1(a)=a$ and $x\_2(a)=a-1$. Moreover, $f\big(x^\*(a),a\big)=g\big(x\_1(a),a\big)$ for $a<0$ and $f\big(x^\*(a),a\big)=h\big(x\_2(a),a\big)$ for $a>0$. As a consequence, the function $a\mapsto x^\*(a)$ jumps down after $0$.

> **Comments:**
> **username_0**: Thanks. Would there be a more definite result if h(x,a) = h(x)?  
> **username_2**: Not really: Let $h(x,a)=-x^2$ and $g(x,a)=a-(a-1-x)^2$.  

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