Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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 $y\gt0$.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.