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Let $W_t$ is standard Brownian motion under probability measure $P$. Consider 1-D stochastic differential equation $$ dY_t = dt + \sigma(Y_t) dW_t, \ Y_0 = y\ge 0.$$ We assume $\sigma(0) = 0$, and $\sigma(x)$ is locally Holder-1/2, i.e. for any bounded subinterval $I\subset [0,\infty)$, we assume there exists constant $K_I$ s.t. $$|\sigma(x_1)- \sigma(x_2)| \le K_I |x_1 - x_2|^{1/2}, \ \forall x_1, x_2 \in I.$$ Note that, the above SDE has strong non-negative solution by comparison with $d X_t = \sigma(X_t) dW_t$.

[Q1] Define $\tau = \inf[t>0: Y_t = 0]$. Is $\tau>0$ almost surely?

[Q2] Can one show that $Y_t > 0$ almost surely for arbitrary given $t>0$?

In fact, it's enough to show the above results with $y= 0$.

My guess is that, [Q2] is too strong to be true, but [Q1] is correct. It will be helpful to get a proof of [Q1] at least.

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why don't you use the scale function ? – mike Aug 2 '12 at 11:46
@Mike, could you be more specific? I have difficulty use scale function, since $\sigma^{-2}$ may not be locally integrable at $0$. I am assuming initial $y = 0$. – kenneth Aug 2 '12 at 13:02
Suppose you divide it into 2 cases, one where $\sigma = 0$ in an interval near $0$, and the other where $\exists x_i \rightarrow 0 $ with $\sigma(x_i) = 0$. If $\ell_i = sup \lbrace x < x_i : \sigma(x) = 0$ then it seems to me that you can show that a process started from $x_i$ never hits $\ell_i$. Then, if the process is positive with probability 1 at time t it would postive with probability 1 for all t. – mike Aug 2 '12 at 15:59
@kenneth: You can answer Q1 by comparing with a Bessel squared process, and the answer depends on the value of $K_I$. On the other hand, Q2 doesn't look very strong to me. I expect that the answer is an unconditional yes. – George Lowther Aug 2 '12 at 17:00
To continue George's comment, the exact probabilities of hitting zero for the BESQ-process are given e.g. in Delbaen and Shirakawa, A Note of Option Pricing for Constant Elasticity of Variance Model, – Stephan Sturm Aug 2 '12 at 17:13

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