# Absolute value of Brownian motion

I need to show that $$R_t=\frac{1}{|B_t|}$$ is bounded in $\mathcal{L^2}$ for $(t \ge 1)$, where $B_t$ is a 3-dimensional standard Brownian motion.

I am trying to find a bound for $\mathbb{E}[\int_{t=1}^{\infty}R^2_t]$.

Asymptotically $B_t^i$ is between $\sqrt{t}$ and $t$. I also know that $|B_t| \to \infty$, but the rate is not clear.

Hints would be helpful.

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## 1 Answer

Since $B_t$ and $\sqrt{t}B_1$ are identically distributed, $\mathrm E(R_t^2)=t^{-1}\mathrm E(R_1^2)$, hence $\mathrm E(R_t^2)\leqslant\mathrm E(R_1^2)$ for every $t\geqslant1$ and it remains to show that $\mathrm E(R_1^2)$ is finite. Now, the density of the distribution of $B_1$ is proportional to $\mathrm e^{-\|x\|^2/2}$ and $B_1$ has dimension $3$ hence the density of the distribution of $Y=\|B_1\|$ is proportional to $\varphi(y)=y^{3-1}\mathrm e^{-y^2/2}=y^2\mathrm e^{-y^2/2}$ on $y\gt0$. Since the function $y\mapsto y^{-2}\varphi(y)=\mathrm e^{-y^2/2}$ is Lebesgue integrable, the random variable $Y^{-2}=R_1^2$ is integrable.

On the other hand, $\mathrm E\left(\int\limits_1^{+\infty}R_t^2\mathrm dt\right)$ is infinite.

Edit The distribution of $B_1$ yields the distribution of $Y=\|B_1\|$ by the usual change of variables technique. To see this, note that in dimension $n$ and for every test function $u$, $$\mathrm E(u(Y))\propto\int_{\mathbb R^n} u(\|x\|)\mathrm e^{-\|x\|^2/2}\mathrm dx\propto\int_0^{+\infty}\int_{S^{n-1}}u(y)\mathrm e^{-y^2/2}y^{n-1}\mathrm d\sigma_{n-1}(\theta)\mathrm dy,$$ where $\sigma_{n-1}$ denotes the uniform distribution on the unit sphere $S^{n-1}$ and $(y,\theta)\mapsto y^{n-1}$ is proportional to the Jacobian of the transformation $x\mapsto(y,\theta)$ from $\mathbb R^n\setminus\{0\}$ to $\mathbb R_+^*\times S^{n-1}$. Hence, $$\mathrm E(u(Y))\propto\int_0^{+\infty}u(y)y^{n-1}\mathrm e^{-y^2/2}\mathrm dy,$$ which proves by identification that the distribution of $Y$ has a density proportional to $y^{n-1}\mathrm e^{-y^2/2}$ on $y\gt0$.

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Firstly I should note that I was asked about $\mathcal{L}^2(\Omega,\mathbb{P})$ as opposed to $\mathcal{L}^2(\Omega \times\mathbb{R},\mathbb{P\times\lambda})$. Could you expand on how you calculated density of $BES_0(3)=Y$ process? – Tom Artiom Fiodorov Jan 16 '12 at 21:34
I was asked about $\mathcal{L}^2(\Omega,\mathbb{P})$ as opposed to $\mathcal{L}^2(\Omega \times\mathbb{R},\mathbb{P\times\lambda})$... ?? – Did Jan 17 '12 at 7:13
Thank you very much for you incredible input. I am still trying to figure out the transformation. Didn't see $d\sigma$ before. By $\mathcal{L}^2$ I meant to say that you calculated $\mathbb{E}[R_t^2]$, I was doing $\mathbb{E}[\int{R_t^2dt}]$. – Tom Artiom Fiodorov Jan 18 '12 at 17:11
These are sometimes called hyperspherical coordinates. // Yes, the post shows $E(R_t^2)$ is finite and how to compute its value, and shows that $\int E(R_t^2)dt=E(\int R_t^2dt)$ is infinite. – Did Jan 18 '12 at 18:07
I'd love see a proper answer to the true question conserning the bound to $\mathbb{B} \left\{\int_1 ^{\infty} R_s^2ds \right \}$ – Paul Jan 3 '13 at 10:28