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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?

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1 Answer 1

$W_t$ 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 $Y_t$.

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@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

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