skewness of a truncated distribution function Let $Y$ be a real random variable and $a$ a real number. Let $X:=\min(a,Y)$. Is it true that the skewness of $X$ is an increasing function of $a$?
 A: Given translation-invariance, we can ask equivalently if, when
$X_a := \min(0, Y-a)$ and $\gamma(a)$ is the skewness of $X_a$,
$\gamma$ is increasing. (I assume that by increasing you don't
mean strictly increasing)
I think this is true and will attempt to sketch a proof for the
case that $Y$ has a p.d.f. $f$.
The moments of $X_a$ can in this case be found by
$$
M_n(a) = \int_{-\infty}^a (x-a)^n f(x) dx.
$$
By Leibniz's rule, we find that
$$
M_n'(a) = (a-a)^n f(a) + \int_{-\infty}^a -n(x-a)^{n-1} f(x) dx = -n M_{n-1} (a)
$$
for $n >= 2$, and
$$
M_1'(a) = (a-a)f(a) + \int_{-\infty}^a -f(x) dx = -F(a)
$$
where $F$ is the c.d.f. of $Y$.
We can then express the skewness as 
$$
\gamma = (M_3 - 3 M_1 M_2 + 2 M_1^3) (M_2 - M_1^2)^{-3/2}
$$
and differentiate to get
\begin{eqnarray*}
\gamma' &=& (M_3' - 3 M_1' M_2 - 3 M_1 M_2' + 6 M_1^2M_1') (M_2 - M_1^2)^{-3/2} \\\\
    & &+ (M_3 - 3 M_1 M_2 + 2M_1^3) (-\frac{3}{2}) (M_2 - M_1^2)^{-5/2} (M_2' - 2 M_1 M_1') \\\\
  &=& (-3M_2 + 3F M_2 + 6 M_1^2 - 6 M_1^2 F) (M_2 - M_1^2)^{-3/2} \\\\
    & &+ (M_3 - 3 M_1 M_2 + 2 M_1^3)(-\frac{3}{2})(M_2 - M_1^2)^{-5/2}(-2M_1 +2M_1 F) \\\\
  &=& 3 (1 - F) (M_2 - M_1^2)^{-5/2} (M_1 M_3 - M_2^2)
\end{eqnarray*}
Since the $X_a$ are all non-positive, we can apply Cauchy-Schwarz to
$(-X_a)^{1/2}$ and $(-X_a)^{3/2}$ to find $M_1 M_3 \ge M_2^2$, showing that
$\gamma' \ge 0$.
I think this argument can be extended to arbitrary distributions by
using one-sided derivatives. The main obstacle is that $M_1$ is not differentiable
at point masses, but the right derivative still exists and is as above.
