Distribution of $X_t=\mathbb{E}(B_1\mathbb{I}_{|B_1|\geq1}-B_1\mathbb{I}_{|B_1|<1}|\mathcal{F}_t)$? I am asked to consider the process
$$X_t=\mathbb{E}(X|\mathcal{F}_t),\quad X=B_1\mathbb{I}_{|B_1|\geq1}-B_1\mathbb{I}_{|B_1|<1}=\begin{cases}B_1&\text{if }|B_1|\geq1\\-B_1&\text{otherwise}\end{cases}$$
for $t\in[0,1]$, where $\{B_t\}$ is a standard BM and $\mathcal{F}_t=\sigma(B_s:s\in[0,t])$. I am tasked to obtain the following:
i) Find the distribution of $X$ and $X_t$, and prove $X_t$ is a martingale;
ii) Show that $X_t=B_t-2f(B_t,\sqrt{1-t})$ where $f(x,s)=\mathbb{E}\left[(x+sZ)\mathbb{I}_{|x+sZ|<1}\right]$ and $Z\sim\mathcal{N}(0,1)$, and consequently derive a PDE with an initial condition based on $s$.
For i), the comments and current answer have explained why it is a martingale. I also now know that for ii), we have to use Ito’s formula and make the drift term disappear.
Any guidance is greatly appreciated!
 A: This is not a full answer but it may be a hint to further solution.
$X_t$ is Levy martingale hence it's martingale.
Put $\xi = B_t$ and $\eta = B_1 - B_t$.
Thus $X_t = E( B_1 - 2 B_1 I_{|B_1| < 1} | \mathcal{F}_t) = E(h(\xi, 
\eta) | \mathcal{F}_t) $
where $h(a,b) = a + b- 2(a +b) I_{|a + b| < 1}$.
We know that $\xi$ is $\mathcal{F}_t$-measurable. As $\eta$ and $\mathcal{F}_t$ are independent we have
$$X_t = E(h(\xi, \eta) | \mathcal{F}_t) = g(\xi),$$
where $g(x) =  E(h(\xi, \eta) | \xi = x)$.
Further, $g(x) =  E(h(\xi, \eta) | \xi = x) =  E(h(x, \eta) | \xi = x) = Eh(x, \eta) $ by independence. Put $\eta = \sqrt{1-t} \zeta$. As $\eta \sim N(0, 1-t)$ we have $\zeta \sim N(0,1)$. Thus
$$g(x) = Eh(x, \eta) = E( x + \eta- 2(x +\eta) I_{|x + \eta| < 1} )= x - 2E(x +\eta) I_{|x + \eta| < 1} $$
$$= x -  2E(x +\sqrt{1-t} \zeta) I_{|x +\sqrt{1-t} \zeta| < 1} = x - 2f(x, \sqrt{1-t})$$
and
$$X_t = g(\xi) =  \xi - 2f(\xi, \sqrt{1-t}) = B_t - 2f(B_t, \sqrt{1-t}) .$$
We proved that $X_t = H(B_t, t)$. Now we may try to find the distribution of $X_t$ or use Ito formula, but a function $f(x,s)$ doesn't look good.
