# Show that semimartingale related to Brownian motion is local martingale

I got stuck on another problem regarding whether certain process is a local martingale.

Set $T_\epsilon=\inf\{ t\ge 0:B_t=\epsilon\}$, $\lambda >0$ and $\alpha\neq 0$ Show that the process $$Z_t=(B_{t\wedge T_\epsilon})^\alpha\exp\left(-\lambda \int_0^{t\wedge T_\epsilon}\frac{ds}{B_s^2}\right)$$ is a local martingale if $\alpha$ and $\lambda$ satisfy an polynomial equation.

As a first step, I applied Ito's lemma to $Z_t$ which gives us $$Z_t=Z_0+\alpha\int_0^{t\wedge T_\epsilon}B_s^{\alpha-1}dB_s-\lambda\int_0^{t\wedge T_\epsilon}\exp(-\lambda Y_s)dY_s+$$ $$+\frac{1}{2}\left[\frac{\alpha(\alpha-1)}{2}\int_0^{t\wedge T_\epsilon}B_s^{\alpha-2}ds +\lambda^2\int_0^{t\wedge T_\epsilon}\exp(-\lambda Y_s)d\langle Y,Y\rangle_s+2(\ldots)\int_0^{t\wedge T_\epsilon}d\langle B,Y\rangle_s\right]$$ However during the computation of $\langle Y,Y\rangle_s$ I met some problems: quadratic variation $\langle Y,Y\rangle_s$ is the limit in probability of $M^{(n)}:=\sum_{i=1}^{p_n}\left(\int_{t_{i-1}}^{t_i}\frac{ds}{B_s^2}\right)^2$, where subdivision's mesh $\sup_{i\le p_n}|t_i-t_{i-1}|\to 0$. So, I wanted to find the moments of $\int_{t_{i-1}}^{t_i}\frac{ds}{B_s^2}$ to see where $M^{(n)}$ converge to as $n\to \infty$. But apparently $$\mathbb{E}\int_{t_{i-1}}^{t_i}\frac{ds}{B_s^2}=\int_{t_{i-1}}^{t_i}\mathbb{E}\frac{1}{B_s^2}ds$$ and simple computations show that the last expectation doesn't exist as for normally distributed random variable $\xi$ the expectation $\mathbb{E}\frac{1}{\xi^2}$ is infinite.

• If $B_0>\varepsilon$, $Y_t=\int_0^{t\wedge T_{\varepsilon}}\frac{ds}{B_s^2}$ is a process with finite variation, and $\langle Y,Y\rangle=0$, $\langle B,Y\rangle=0$. – JGWang Jun 4 '17 at 2:58
• Many thanks! Assume that $B_0=1$. I am wondering whether $Y_t$ is of finite variation because the variation of $Y_t$ over $[0,T]$ is bounded due to $\sup \sum_{k}\left|Y_{t_k}-Y_{t_{k-1}}\right|=\sup \sum_{k}\left|\int_{t_{k-1}}^{t_{k}}\frac{ds\wedge T_\epsilon}{B_{s\wedge T_\epsilon}^2}\right|\le \frac{1}{\epsilon^2}\sum_{k}|t_k-t_{k-1}|= \frac{T}{\epsilon^2}<\infty$. And as any finite variation process its quadratic variation $\langle Y,Y\rangle_t=0$ – Mushtandoid Jun 4 '17 at 13:34