MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The squared Bessel process with $\delta$-dimension for $\delta>0$, denoted by $BESQ^\delta(y)$, is given by $$d Y_t = \delta t + 2 \sqrt{Y_t} d B_t, \ Y_0 = y\ge 0$$ where $B_t$ is BM under $(\Omega, {\cal F}_t, P)$. Consider $\tau = \inf[ t>0: Y_t = 0].$

(Claim). $\tau = \infty$ almost surely.

(Proof). Let $X_t = Y_{\frac 2 \delta t}$. Then, $$d X_t = 2 t + 2 \sqrt{X_t} d W_t, \ X_0 = y,$$ where $W_t = B_{\frac 2 \delta t}$ is BM under $(\Omega, {\cal F}_{\frac 2 \delta t}, P)$. In other words, $X_t$ is $BESQ^2(y)$ w.r.t. time-scaled filtration under the same probability measure. Therefore, {0} is polar set of $X_t$, so is of $Y_t$. END.

However, it gives a contradiction to the fact that $\tau = 0$ for $BESQ^1(0)$ due to the properties of 1-D BM. Where is the gap of the above proof?

share|cite|improve this question

$W_t$ is not a Brownian motion, you need to rescale it and rather use $V_t = \sqrt{\frac{\delta}{2}} B_{\frac{2}{\delta}t}$. If you denote $X_t' = \frac{\delta}{2} X_t$, then you will see that it satisfies the same equation as $Y_t$.

share|cite|improve this answer
@Mateusz, Thanks for your answer – kenneth Aug 3 '12 at 12:04
@kenneth: You're welcome; I hope everything is clear now. – Mateusz Wasilewski Aug 3 '12 at 12:38

Your Answer


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.