The minimizer of $E\left[\left|I\left(V \leq v\right)-a\right|^{\alpha}\right]$ Why the minimizer of $E\left[\left|I\left(V \leq v\right)-a\right|^{\alpha}\right]$ w.r.t. $a$ is given by
$$
Q_{V}^{\alpha}(v)=\frac{1}{1+\left\{\frac{F_{V}(v)}{1-F_{V}(v)}\right\}^{1 /(1-\alpha)}}
$$
where $F_{V}(v)$ is cumulative distribution function of random variable $V$, and $\alpha$ is a fixed constant.
 A: Let us call the objective function
$$
f(a) = \text{E}[|I(V\leqslant v) - a|^\alpha] 
     = |1-a|^\alpha F_V(v)+|a|^\alpha (1-F_V(v)) 
$$
and find the extremal points for $a$ by differentiating this function to $a$ and check for which $a$ we get that $f'(a)=0$. We can get rid of the absolute values by considering the function $f(a)$ in appropriate intervals:
$$
f(a) = \begin{cases}(1-a)^\alpha F_V + (-a)^\alpha(1-F_V) &, a<0\\ 
                    (1-a)^\alpha F_V +  a^\alpha(1-F_V) &, 0\leqslant a\leqslant 1\\
                    (a-1)^\alpha F_V + a^\alpha(1-F_V) &, a>1 
                    \end{cases} 
$$
Now, it can be verified that $f'(a)<0$ when $a<0$ and that $f'(a)>0$ when $a>1$, so any extrema are to be found in $0\leqslant a \leqslant 1$. Imposing $f'(a)=0$ in that region gives
$$
  -\alpha(1-a)^{\alpha-1} + \alpha a^{\alpha-1}(1-F_V) = 0
$$
which solves to
$$
  a = \frac{1}{\displaystyle 1+\Big(\frac{1-F_V}{F_V}\Big)^{\frac{1}{\alpha-1}}}
    = \frac{1}{\displaystyle 1+\Big(\frac{F_V}{1-F_V}\Big)^{\frac{1}{1-\alpha}}}
$$
