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I would like to know if my understanding about how to find a maximum of the function when some parameters are unknown is correct.

Consider the following maximization problem.

$\max_{x}V=\int_0^{a(x)}y(x)dF(t)+\int_{a(x)}^{\infty}z(x)dF(t)$

where $x\geq0$, $y'(x)>0$, and $z'(x)>0.$ If $x\to\infty$, $z(x)\to \overline{z}$. $F$ is the cumulative distribution function of $t$.

The function $a(x)$ is non-monotonic in $x$:

if $x=0$, $a(x)=0$,

if $0<x<\infty$, $a(x)>0$,

and if $x\to\infty$, $a(x)=0$.

In this maximization problem, if we increase $x$, the first term in the right side first increases and then decreases to zero because of the nonmonotonicity of $a(x)$.

If $y(x^*)+z(x^*)> y(x) +z(x)$ for any $x$, the objective function $V$ is maximized at $x=x^*$, and the maximum value is $\int_0^{a(x^*)}y(x^*)dF(t)+\int_{a(x^*)}^{\infty}z(x^*)dF(t)$.

If $y(x)+z(x)<\overline{z}$ for any $x$, the objective function $V$ is maximized at $x=\infty$, and the maximum value is $\int_{0}^{\infty}\overline{z}dF(t)$. The first term in the right side is zero.

I would appreciate it if someone could tell me if my understanding is correct. If it is incorrect, where did I make a mistake?

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  • $\begingroup$ There seems to be a problem with your variable $t$ of integration which does not appear as argument of $y$ and $z$. $\endgroup$ – Bertrand May 23 at 18:54
  • $\begingroup$ did you mean to say $dF(x)$? $\endgroup$ – Regio May 23 at 19:00
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Disclaimer: I'm assuming you are integrating over $x$:

I have a problem with your first inequality. Since $y(x)$ and $z(x)$ are monotonically increasing (strictly) with respect to $x$, there cannot exists an $x^*$ such that $y(x^*)+z(x^*)>y(x)+z(x)$ for all $x\geq 0$. So the statement is vacuously true.

The second observation seems correct to me.

In general, you could use the Leibniz rule to study the FOC's assuming the solution is interior. Note that your first inequality is not necessary for an interior solution. So the fact that it cannot occur does not mean there cannot be an interior solution.

Note as well that even if $y(x)>\bar z$ for some values of $x$, the solution can still be at $x=\infty$ (assuming you can choose such a value of $x$, of course).

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  • $\begingroup$ Thank you so much for your reply. $\endgroup$ – Tom M. May 23 at 19:32

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