Show that the expected absolute deviation is quadratic near 0 Let $X$ be a real valued random variable with median $0$ having a continuous, positive density (with respect to lebesgue measure) $f$ in a neighborhood of $0$. Show that:
$$g(t) = \mathbb E(|X - t| - |X|) = t^2f(0) + o(t^2)$$
as $t\rightarrow 0$.
I'm not sure how to show this since I would usually apply a taylor expansion for this kind of result, but my quantity of interest, $g$, appears to be non-differentiable with respect to $t$. If I could compute the expectation or approximate it somehow, then I might be able to show the result.
EDIT: for context, this is the first claim in section 2 of Pollard's Asymptotics for least absolute deviation regression estimators. See link.
 A: Let $P_X$ be the measure induced by $X$. We are given that $P_X$ has a continuous, positive density $f$ in a neighbourhood of $0$.
Since $0$ is a (the) median of $X$, the conditions on $f$ allow us to conclude $P_X (-\infty,0] = P_X [0,\infty) = {1 \over 2}$.
Let $\epsilon>0$ and choose $\delta>0$ such that $f$ is defined and positive on $[-\delta,\delta]$ and $|f(0)-f(\xi)| < \epsilon$ for $|\xi| < \delta$.
Choose $t \in [0,\delta)$, then
\begin{eqnarray}
E[|X-t|-X] &=& \int (|\xi-t| - |\xi|) P_X(d \xi) \\
&=& \int_{-\infty}^0 t P_X(d \xi) + \int_0^t (t-2 \xi) P_X(d \xi) - \int_t^\infty t P_X(d \xi) \\
&=&  \int_0^t (t-2 \xi) P_X(d \xi) + \int_0^t t P_X(d \xi) \\
&=& 2\int_0^t (t-\xi) f(\xi) m(d \xi) \\
&=& 2\int_0^t (t-\xi) f(0) m(d \xi) + 2\int_0^t (t-\xi) (f(\xi)-f(0)) m(d \xi)\\
&=& t^2 f(0) + 2\int_0^t (t-\xi) (f(\xi)-f(0)) m(d \xi)\\
\end{eqnarray}
The same analysis applies mutatis mutandis for $t \in (-\delta,0]$.
Under the above assumptions we see that
$|2\int_0^t (t-\xi) (f(\xi)-f(0)) m(d \xi)|  \le \epsilon t^2$.
