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I am going through the exercises of Brownian Motion and Stochastic Calculus by Karatzas, and I have been stuck for a while trying to prove the following result.

Assuming $R$ is a Bessel process with dimension $d \geq 3$, starting at $r = 0$, I would like to show that the process $\{M_t = (1 / R_t^{\,d-2}); 1 \leq t < \infty \}$

  1. is a local martingale, and
  2. is not a true martingale.

Whenever I come across proving results that involve local martingales, it always confuses me their very definition (involving a sequence of stopping times) and find it hard to operate with them. I'd be very glad if anybody can shed some light into the matter.

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1 Answer 1

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I wanted to provide some details on the proof, please let me know any further details are required. To prove that $M_t$ is a local martingale consider the following stopping times: $T_k=\inf\{t: R_t=\frac 1k\}$, $S_k=\inf\{t: R_t=k\}$, $\tau_k=T_k\wedge S_k\wedge n$. We know that $R_t$ satisfies $R_t=\int_{0^t}\frac{d-1}{2R_s}ds + B_t$, where $B_t$ is a Brownian Motion (see Karatzas, Shreve, proposition III.3.21). We can note that stopped processes $B_{t\wedge \tau_k}$ take values within $[1/k,k]$ where function $f$ is twice continuously differentable, thus allowing to apply Ito formula for $f(R_t)$. Doing this, because of the choice of the function $f(x)=x^{2-d}$, we see that $f(R_{t\wedge \tau_k})$ will be a local martingale satisfying $$ M_{t\wedge \tau_k}=M_{1\wedge \tau_k}+(2-d)\int_{1}^{t\wedge \tau_k}R_s^{1-d}dB_s $$ Using the fact that $\tau_k$ is bounded stopping time (because Bessel process of dimension $d\ge 3$ almost surely doesn't reach the origin and $R_t\to \infty$ $(t\to \infty)$) and boundedness of $R_s$ in $s\in (1,\tau_k)$ we conclude that $\int_{1}^{t\wedge \tau_k}R_s^{1-d}dB_s$ is a martingale. Using these arguments we also have that $\tau_k\uparrow \infty$.

Next, we can show that in dimension $d=3$, $M_t=R_t^{-1}$ is not a martingale by computing $\mathbb{E}(M_t)$. $$ \mathbb{E}(M_t)=\int_{\mathbb{R}^3}\frac{1}{\sqrt{x^2+y^2+z^2}}\frac{1}{(\sqrt{2\pi t})^3}\exp\left(-\frac{x^2+y^2+z^2}{2t}\right)dxdydz $$ Rewriting this integral in spherical coordinates $(x,y,z)\mapsto (r, \phi, \theta)$, we can rewrite it as $$ \frac{1}{(\sqrt{2\pi t})^3}\int_{0}^\infty\int_{0}^{2\pi}\int_{0}^{\pi}\frac 1r\exp(-\frac{r^2}{2t})r^2\sin\theta drd\phi d\theta= $$ $$ =\frac{1}{(\sqrt{2\pi t})^3}\int_{0}^\infty\frac 1r\exp(-\frac{r^2}{2t})r^2 dr\cdot 2\pi \int_{0}^{\pi}\sin \theta d\theta= $$ $$ =\frac{2\cdot 2\pi t}{(\sqrt{2\pi t})^3}\int_{0}^\infty\exp(-u) du =\sqrt{\frac{2}{\pi t}} $$ If $M_t$ were martingale then $\mathbb{E}(M_t)$ had to be the same for all $t\ge 1$, but above computation show that this expectation depends on $t$.

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