Differentiating $G$ with respect to $c$ and setting this derivative equal to zero gives
$$
\begin{aligned}
\frac{d G(c)}{d c} &=[1-F(c)]-c f(c)=0 \
\text { i.e., } \quad \frac{f(c)}{1-F(c)} &=\frac{1}{c}
\end{aligned}
$$
b. The left-hand side of the last equation is the hazard function $h$ of $\mathbf{U}$; cf. page 2. The equation states that the optimal strategy $c$ is the abscissa at which the hazard function $h(c)$ crosses the function $1 / c$.
c. For an exponential rv, the hazard function is constant, $h(c)=\lambda$. Therefore, the optimal strategy is $c=1 / \lambda$, i.e., to choose the expected value of $\mathbf{U}$, With this strategy, the expected gain is
$$
G\left(\frac{1}{\lambda}\right)=\frac{1}{\lambda} \cdot e^{-1}
$$
i.e., about $36.8 \%$ of the expected value of $\mathbf{U}$. This value appears remarkably small, but due to the large spread (relative to its mean) of the exponential distribution, there is no way to further improve it within the given setting.
STAT4028 COURSE NOTES ：
With the strategy $c$, the gain (actually, loss) will be equal to $-c$ with probability $\mathrm{P}(\mathbf{U} \leq c)=F(c)$, and it will be equal to $+c$ with probability $\mathrm{P}(\mathbf{U}>c)=1-F(c)$. Weighing these two cases by their respective probabilities, we get
$$
\begin{aligned}
G(c) &=-c \cdot F(c)+c \cdot[1-F(c)] \
&=c \cdot[1-2 F(c)]
\end{aligned}
$$
Figure $3.6$ shows the qualitative behavior of $G$.
Differentiating $G$ with respect to $c$ and setting this derivative equal to zero yields the optimum condition
$$
\frac{d G(c)}{d c}=1-2[c f(c)+F(c)]=0
$$