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Bessel process of dimension 2 is defined to be solution of $$ dX_t=dB_t+\frac{1}{2X_t}dt,\quad X_0=x_0>0 $$ where $B$ is a standard 1-dimensional Brownian motion. $X$ can be viewed as the norm of a standard 2-dimensional Brownian motion. It returns arbitrarily close to $0$ but will never hit $0$.

It is known that for any $\epsilon>0$, Bessel process of dimension $2-\epsilon$ which is defined to be solution of $$ dX_t=dB_t+\frac{1-\epsilon}{2X_t}dt,\quad X_0=x_0>0 $$ almost surely hits $0$.

Now let $f$ be a certain function such that $f(y)\to 0$ and $f(y)/y\to\infty$ as $y\to 0$. Let $Y_t$ be a solution to $$ dY_t=dB_t+\frac{1-f(Y_t)}{2Y_t}dt. $$ Is there a criterion to judge if $Y_t$ will hit $0$? For example, what will happen when $f(y)=y^c$ for $0<c<1$? What about $f(y)=-y\log y$?

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As usual this kind of questions can be answered using the Feller test for explosion (see for instance Karatzas, Shreve). Before doing the computations, let me notice that $f\equiv0$ is the Bessel process, while $f\equiv\epsilon$ is the example given. Therefore, the strength of the order of zero of $f$ should be the main parameter. The stronger the zero, the closer to Bessel is the process $Y$; the weaker, the closer to the $\epsilon$ case. In other words, conditions on $f(y)/y$ should not be sufficient (and indeed they aren't, see below).

Now the computations. Let $$ G(x) = \operatorname{e}^{-\int_x^1\frac{f(z)}{z}\,dz}, $$ then the scale function $\varphi$ and the speed measure $m$ for $Y$ are given by $$ \varphi(x)=-\int_x^1\frac{G(y)}{y}\,dy, \qquad m(x)=\frac{x}{G(x)}. $$ The Feller test yields,

  1. if $\int_x^1\frac{G(y)}{y}\,dy=\infty$, then $\mathbb{P}[\tau_0=\infty]=1$,
  2. if $\int_x^1\frac{G(y)}{y}\,dy<\infty$, then $\mathbb{P}[\tau_0=\infty]=1$ if and only if $\int_0^1 (\varphi(x)-\varphi(0))m(x)\,dx=\infty$,

where $\tau_0$ is the hitting time of $0$.

Clearly, (1) happens for $f(x)=x^c$, $0<c<1$ and $f(x)=-x\log x$ (the order of zero in $0$ is `too' strong). To see an instance of (2) with positive probability of hitting zero, take $f(x)=\frac{a}{-\log x}$ with $a>1$.

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