For a Brownian motion $(B_t)_{t\geq0}$ we define the process
$$I_t:=\int_0^t\frac{|B_s|}{s}\Bbb dB_s,\qquad (t>0).$$
But what can we say about this process? Is it even possible to define this process? If yes, it is for sure a local martingale, but is it also a (true) martingale? As
$$\Bbb E[I]_t=\Bbb E\int_0^t\frac{B_s^2}{s^2}\Bbb ds=\int_0^t\frac{1}{s}\Bbb ds=\infty,
$$
where $[.]_t$ denotes the quadratic variation, the second moment can't exist if it is a martingale. Also, what can we say about the limit $$I_0:=\lim\limits_{t\searrow0}I_t,$$ does it exist in any sense?
I thought about scaling property, Hölder continuity and the law of the iterated logarithm, but notthing helped. As the limit $$\lim\limits_{t\searrow0}\frac{B_t^2}{t}$$ doesn't exist, I can't just use the Itô formula.
All hints about which properties this process possesses are welcome.
Best Answer
Consider $$ I_t = \int_{t}^{2t} \frac{|B_s|}{s} dB_s = \int_1^2 \frac{|B_{tu}|}{tu} dB_{tu}. $$ By the self-similarity of $B$, $\{B_{tu}, u\in [1,2]\}$ has the same distribution as $\{\sqrt{t}B_{u}, u\in [1,2]\}$. Therefore, $I_t$ has the same law$^*$ as $$ \int_1^2 \frac{\sqrt{t}|B_{u}|}{tu} \sqrt{t} \,dB_{u} = \int_1^2 \frac{|B_{u}|}{u} dB_{u} = I_1. $$ On the other hand, if the integral in question were well defined, then we would have $I_t\to 0, t\to0+$, which is impossible (as $I_1$ is not zero a.s.).
$^*$ I believe this can be shown in many ways, here's one of the arguments.
Let $B$ be defined canonically, i.e., $\Omega = \{\omega\in C([0,\infty)): \omega(0) = 0\}$, $\mathcal F = \mathcal B(C([0,\infty))$, $\mathrm P$ is the Wiener measure, $B(t,\omega) = \omega(t)$; $\mathcal F_t = \sigma(\{\omega(s) \in A\}, s\in[0,t], A\in \mathcal B(\mathbb R)\}$.
The self-similarity means that, for each $C>0$, the transformation $F_C: \omega(\cdot) \mapsto C^{-1/2}\omega(C\cdot)$ preserves $\mathrm P$. Moreover, it agrees with the filtration $(\mathcal F_t,t\ge 0)$ in the sense that $F_C(\mathcal F_t) = \mathcal F_{Ct}, t\ge 0$.
Consider now the Itô integral $I(\xi,B)=\int_0^1 \xi_s dB_s$. It is obtained as a continuous extension (first in $L^2$ for isometric definition, then in probability for extended definition) of integral of simple processes. As a result, $I(\xi,B)(F_C(\omega)) = I\bigl(\xi(F_C(\omega)),B(F_C(\omega))\bigr)$ a.s.: this equality is obvious for simple processes, and then is extended by continuity. On the other hand, since $F_C$ perserves $\mathrm P$, $I(\xi,B)(F_C(\omega))$ and $I(\xi,B)(\omega)$ have the same distribution.
In particular, for $\xi_t = f(t,B_t)$, we have $$ I(\xi,B)(F_C(\omega)) = \int_0^1 f(s,C^{-1/2}B_{Cs}) C^{-1/2} dB_{Cs}\overset{d}= I(\xi,B) = \int_0^1 f(s,B_{s}) dB_{s}. $$