I want to solve the following Constrained optimization program:
$\begin{equation} \begin{split} \max_{\{F\}} F \cdot P\left(v \geq -\alpha \cdot \ln\left(1-\frac{F}{\alpha}\right)\right)\\ \text{Subject to: } F>0, F<\alpha \end{split} \end{equation}$
Where $\alpha$ is a (real) parameter strictly greater than $0$, i.e $\alpha \in {R}^*$, and where $v$ follows a (continuous) Uniform distribution over (0,1).
I wish to find a closed form solution for $F$, as a function of $\alpha$.
Here is what I have done so far:
The above program is equivalent to: \begin{equation} \begin{split} \max_{\{F\}} F \cdot \left[1-P\left(v \leq -\alpha \cdot \ln\left(1-\frac{F}{\alpha}\right)\right)\right] \\ \text{Subject to: } F>0, F<\alpha\\ \end{split} \end{equation}
Then, we know: $$P\left(v \leq -\alpha \cdot \ln\left(1-\frac{F}{\alpha}\right)\right) = \begin{cases} 0 &\text{ If} -\alpha \cdot \ln(1-\frac{F}{\alpha}) < 0,\\ -\alpha \cdot \ln(1-\frac{F}{\alpha}) &\text{ If} -\alpha \cdot \ln(1-\frac{F}{\alpha}) \in [0,1],\\ 1 &\text{ If} -\alpha \cdot \ln(1-\frac{F}{\alpha}) > 1 \end{cases} $$
But: (a) Having $F>0$ (one of the constraint), ensures that $-\alpha \cdot \ln(1-\frac{F}{\alpha}) > 0$, since $\alpha > 0$. Hence the first region of the CDF (when $<0$) can be ignored. And, (b) We see clearly that $F$ such that $-\alpha \cdot ln(1-\frac{F}{\alpha}) > 1$ is never optimal, as the objective function would be equal to zero in that case: $ F \cdot \bigg[1- 1 \bigg] = 0$
Hence, the previous program is in turn equivalent to: \begin{equation} \begin{split} \max_{\{F\}} F \cdot \bigg[1- \big( -\alpha \cdot \ln(1-\frac{F}{\alpha}) \big) \bigg] = \underbrace{F \cdot \bigg[1+\alpha \cdot \ln(1-\frac{F}{\alpha}) \bigg]}_{\stackrel{def}{=}f(F)} \\ \text{Subject to: } F>0, F<\alpha, F \leq \alpha (1-e^{-1/\alpha})\\ \end{split} \end{equation} Where the added third condition ensures that $-\alpha \cdot \ln(1-\frac{F}{\alpha}) \in (0,1]$
The first order condition is: \begin{equation} \begin{split} \frac{\partial f(F)}{\partial F}: 1 + \alpha \cdot \ln\left(1-\frac{F}{\alpha}\right) - \frac{\alpha F}{\alpha - F} \stackrel{!}{=} 0 \end{split} \end{equation} Which I don't know how to factorise, and hence solve for F.
I am pretty sure a (unique?) solution exist to problem, as I have plotted the function (and solved it numerically with Mathematica).
Nota Bene: If needed, some restriction on the upper bound of $\alpha$ can be made.
Thank you,
The first order condition is equivalent to
$$1=\exp\bigg[\frac{F(1+\frac{1}{\alpha})-1}{\alpha-F}\bigg]+\frac{F}{\alpha} \quad \quad (1)$$
The function on the RHS takes the values $e^{-1}$ and $1+\frac{1}{\alpha+1}$ at $F=0$ and $F=\frac{1}{1+1/\alpha}$ respectively, and is continuous on $[0,1+\frac{1}{\alpha+1}]$. Therefore, by the intermediate value theorem, there exists $F^*\in(0,\alpha)$ satisfying $(1)$. Since the RHS is strictly increasing on $(0,\alpha)$, there can be only one such $F^*$.
The second derivative of $f(F)$ is
$$\frac{-1}{(1-F/\alpha)}+\frac{-1}{(1-F/\alpha)^2}<0$$
which shows that $f$ is strictly concave on $(0,\alpha)$. Therefore $F^*$ is indeed the unique global maximum.
From the inequality $e^x\geq 1+x$, and the fact that $F^*<\frac{1}{1+1/\alpha}$, we see that $F^*< \alpha (1-e^{-1/\alpha})$. Hence at the optimum the objective function is strictly positive.