Ito integral over an indicator function I have very little experience with Ito integral so I am stuck with this task:
For an Ito process $P_t$ starting at $0$ I want to compute the Ito-integral
$
\begin{align}
\int_0^t \mathbb{I}_{ \{ P_s \geq 0 \}} dP_s.
\end{align}
$
Intuitively this should be  $0 $ if $P_t \leq 0$ and $P_t$ if $P_t \geq 0$, but I don't really know how to approach this since the non-continuity of the indicator doesn't allow to use Ito's formula. Or am I missing somethin there?
 A: As the comment says, the question makes little sense for a geometric Brownian motion. So let $P$ be just a general Ito process with $P_t = \int_0^t a_s ds + \int_0^t b_s dW_s$.
You want to write the Ito formula for $F(x) = \max(0,x)$ with $F'(x) = \mathbb{I}_{x\ge 0}$:
$$
dF(P_t) = \mathbb{I}_{P_t\ge 0} dP_t + \frac{1}{2} F''(P_t) b_t^2\, dt. 
$$
Now what's $F''(x)$? You are correctly saying it does not exist, but informally $F''(x) = \delta_0(x)$, so
$$
F(P_t) = \int_0^t \mathbb{I}_{P_s\ge 0} dP_s + \frac{1}{2}\int_0^t \delta_0(P_s) b^2_s\, ds.
$$
Fortunately, this makes perfect sense via the Tanaka formula. The term
$$
\ell_t^0(P) := \int_0^t \delta_0(P_s) b^2_s\, ds
$$
is the so-called normalized (or Meyer–Tanaka) local time af $P$ at $0$ (alternatively, it can be written as $\int_0^t b^2_s \, dL_s^0(P)$, where $L^0_t(P) = \int_0^t \delta_0(P_s)ds$ is the usual "physicist's" local time at $0$).
To sum up,
$$
\int_0^t \mathbb{I}_{P_s\ge 0} dP_s = \max(P_t,0) - \frac{1}{2} \ell_t^0(P).
$$
