Show that the inverse norm of Brownian motion in $R^3$ is uniformly integrable Let $B$ be an $(\mathcal{F})_t$ Brownian motion with values in $\mathbb{R}^3$, with $B_0 = x \in \mathbb{R^3}\backslash\{0\}$. Let $M_t := \lVert B_t \rVert ^{-1}$.
Show that $M_t$ is uniformly integrable by showing that for all $\gamma \in (0, 3)$, there exists a constant $C_{\gamma}$ such that for all $t \geq 0$, $\mathbb{E}[M_t^{\gamma}] \leq C_{\gamma} \min(1, t^{-\gamma/2})$
The purpose of the problem is to show that there exists a local martingale which is uniformly integrable but is not a martingale. So far I've shown that $M_t$ is a local martingale, by applying Ito's formula on $f(B_t)$, where $f(x, y, z) = (x^2+y^2+z^2)^{-\frac{1}{2}}$. The integral in the quadratic variation ends up being 0.
Edit: I believe it has to do with using the Burkholder-Davis-Gundy inequality, but I can't figure out how.
 A: I think the easiest way is to just show it directly from integrating.  Since $B_t \sim N(x,tI)$, we have
\begin{align*}
\mathbb{E}[M_t^{\gamma}] &= \frac{1}{\sqrt{(2\pi)^3 t^3}}\int_{\mathbb{R}^3}|y|^{-\gamma} e^{-\frac t2 |y-x|^2}dy \\
&= \frac{1}{\sqrt{(2\pi)^3 t^3}} \left(\int_{y \in B(0,1)}|y|^{-\gamma} e^{-\frac t2 |y-x|^2}dy + \int_{y \not\in B(0,1)}|y|^{-\gamma} e^{-\frac t2 |y-x|^2}dy \right) \\
&\le \frac{1}{\sqrt{(2\pi)^3 t^3}} \left(\int_{y \in B(0,1)}|y|^{-\gamma} dy + \int_{y \not\in B(0,1)} e^{-\frac t2 |y-x|^2}dy \right).
\end{align*}
The second integral is finite because it's the Gaussian integral over a subset of $\mathbb{R}^3$.  For the first integral, using spherical integration (and the fact that $\gamma < 3$) we have
\begin{align*}
\int_{y \in B(0,1)} |y|^{-\gamma}dy &= \int_0^{2\pi} \int_0^\pi \int_0^1 r^{-\gamma} r^2 \sin \theta dr d\theta d\phi \\
&=\frac{1}{3-\gamma} \int_0^{2\pi} \int_0^\pi  \sin \theta d\theta d\phi  \\
&= \frac{4 \pi}{3-\gamma} < \infty,
\end{align*}
so we conclude $\mathbb{E}[M_t^{\gamma}] < \infty$ for all $\gamma \in (0,3)$.  Choosing $\gamma \in (1,3)$ shows $M_t$ is uniformly integrable.
