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 $\frac{dx^∗}{da}\ge 0$?
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1$\begingroup$ 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... $\endgroup$– Jesper HybelCommented Mar 21, 2021 at 0:39
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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 $0$.
answered Mar 21, 2021 at 1:18
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$\begingroup$ Thanks. Would there be a more definite result if h(x,a) = h(x)? $\endgroup$– AdamCommented Mar 21, 2021 at 2:57
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$\begingroup$ Not really: Let $h(x,a)=-x^2$ and $g(x,a)=a-(a-1-x)^2$. $\endgroup$ Commented Mar 21, 2021 at 8:02