A wrong proof of Squared Bessel process

2012-08-03T06:09:21Z

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?

Answer by Mateusz Wasilewski for A wrong proof of Squared Bessel process

2012-08-03T09:57:14Z

$W_t$ 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 $Y_t$.

A wrong proof of Squared Bessel process - MathOverflow

A wrong proof of Squared Bessel process

Answer by Mateusz Wasilewski for A wrong proof of Squared Bessel process