Uniqueness of solution to quantile minimization problem I read here: http://librarum.org/book/11685/31 (p. 51, Ex. 3) that quantiles are solutions to certain minimization problem. Here is the proof: http://www.math.ucla.edu/~tom/MathematicalStatistics/Sec18.pdf.
It isn't stated explicitely that the problem can't have other solutions than those described. So I was wondering whether this is true and how to proove it. I can see it when $P(Z=b) = 0$. 
 A: The answer is, yes, one can prove that there is no other estimator $b$ that
minimizes $E(L(\theta,b)).$
To summarize the problem, we define
$$
L(\theta, a) = 
\begin{cases}
k_1 |\theta - a| & \text{if } a \le \theta, \\
k_2 |\theta - a| & \text{if } a > \theta,
\end{cases}
$$
where $k_1 > 0$ and $k_2 > 0.$
We let $p = k_1/(k_1 + k_2)$
and let $b$ be a $p$th quantile of the random variable $\theta$;
that is, $P(\theta \le b) \ge p$ and  $P(\theta \ge b) \ge 1 - p.$
We want to show that for any value $a$,
$$E(L(\theta,a) - L(\theta,b)) \ge 0.$$
I questioned the proof in http://www.math.ucla.edu/~tom/MathematicalStatistics/Sec18.pdf
until I realized that
$$L(\theta,a) = k_1 |\theta - a| I(\theta \ge a) + k_2 |\theta - a| I(\theta < a)$$
is equivalent to
$$L(\theta,a) = k_1 |\theta - a| I(\theta > a) + k_2 |\theta - a| I(\theta \le a),$$
(because in either case, $|\theta - a| = 0$ when $\theta = a$), 
and in addition, $P(\theta \le b) \ge p$ implies $P(\theta > b) \le 1 - p$
and  $P(\theta \ge b) \ge 1 - p$ implies $P(\theta < b) \le p.$
Then indeed, for the case $a > b$, we can use the second formulation of $L(\theta,a)$
to conclude that
$$E(L(\theta,a) - L(\theta,b)) \ge (a - b)(k_2 P(\theta \le b) - k_1 P(\theta > b)).$$
From $a - b > 0,$ $k_1 > 0,$ $k_2 > 0,$ 
$P(\theta \le b) \ge p,$ $P(\theta > b) \le 1 - p,$ and 
$k_2 p - k_1 (1 - p) = k_2 \frac{k_1}{k_1 + k_2} - k_1 \frac{k_2}{k_1 + k_2} = 0,$
it then follows that
$$E(L(\theta,a) - L(\theta,b)) \ge (a - b)(k_2 p - k_1 (1 - p)) = 0.$$
For the case $b > a$, we use the first formulation of $L(\theta,a)$ to find that
$$E(L(\theta,a) - L(\theta,b)) \ge (b - a)(k_1 P(\theta \ge b) - k_2 P(\theta < b)),$$
which (together with other facts already given) implies that
$$E(L(\theta,a) - L(\theta,b)) \ge (b - a)(k_1 (1 - p) - k_2 p) = 0.$$
None of the steps in the proof assumes either that $P(\theta = b) = 0$ or 
that $P(\theta = b) > 0,$ so the proof is equally valid in either of these two cases. 
An immediate consequence is that if $a$ is also a $p$th quantile of $\theta,$
then 
$$E(L(\theta,a) - L(\theta,b)) = 0.$$
Now suppose $a$ is greater than any $p$th quantile of $\theta,$
that is, $P(\theta \ge a) < 1 - p.$
Again, let $b$ be a $p$th quantile of $\theta,$ so $a > b.$
The exact expected value of $L(\theta,a) - L(\theta,b)$ in this case is
$$\begin{align}
E(L(\theta,a) - L(\theta,b)) 
& =  E(-k_1(a - b) I(\theta \ge a) \\
& \qquad\quad + ((k_1 + k_2)(a - \theta) - k_1(a - b)) I(b < \theta < a) \\
& \qquad\quad + k_2(a - b) I(\theta \le b)) \\
& =  (a - b)(k_2 P(\theta \le b)
    - k_1 P(\theta > b)) \\
& \qquad + (k_1 + k_2) E((a - \theta) I(b < \theta < a)).
\end{align}$$
Either $P(b < \theta < a) = 0$ or $P(b < \theta < a) > 0.$
If $P(b < \theta < a) = 0,$ then
$$\begin{align}
E(L(\theta,a) - L(\theta,b)) 
& =  E(k_2(a - b) I(\theta \le b) - k_1(a - b) I(\theta \ge a)) \\
& =  (a - b)(k_2 P(\theta \le b) - k_1 P(\theta \ge a)) \\
& > (a - b)(k_2 p - k_1 (1 - p)) = 0.
\end{align}$$
But if $P(b < \theta < a) > 0,$ then
$E((a - \theta) I(b < \theta < a)) > 0,$ and
$$\begin{align}
E(L(\theta,a) - L(\theta,b)) 
& =  (a - b)(k_2 P(\theta \le b) - k_1 P(\theta > b)) \\
& \qquad + (k_1 + k_2) E((a - \theta) I(b < \theta < a)) \\
& > (a - b)(k_2 p - k_1 (1 - p)) = 0.
\end{align}$$
So we see that no such value $a$ can minimize $E(L(\theta,a).$
On the other hand, suppose $b > a$, where again $a$ is not a $p$th quantile of $\theta$.
Then $P(\theta \le a) < p.$
The exact expected value of $L(\theta,a) - L(\theta,b)$ is
$$\begin{align}
E(L(\theta,a) - L(\theta,b)) 
& =  E(k_1(b - a) I(\theta \ge b) \\
& \qquad\quad + ((k_1 + k_2)(b - \theta) - k_2(b - a)) I(a < \theta < b) \\
& \qquad\quad - k_2(b - a) I(\theta \le a)) \\
& =  (b - a)(k_1 P(\theta \ge b) - k_2 P(\theta < b)) \\
& \qquad + E((k_1 + k_2)(b - \theta) I(a < \theta < b)).
\end{align}$$
Again, either $P(b < \theta < a) = 0$ or $P(b < \theta < a) > 0,$
and in either case
$$E(L(\theta,a) - L(\theta,b)) > 0.$$
So in fact no value $a$ that is not a $p$th quantile of $\theta$
can minimize $E(L(\theta,a).$
This proof still has not made any assumption about whether $P(\theta = b)$ is positive.
In summary, $a$ minimizes $E(L(\theta,a))$ if and only if
$a$ is a $p$th quantile of $\theta.$
I think that's what you meant by "uniqueness" of the solution.
