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I have Ito integral $X=\int_0^T f(t) dW(t)$ and I would like to proof that $P(X>K)>0$ for all $K$ provided $f(t) > \epsilon > 0$.
My idea was $\int_0^T f(t) dW(t) \sim \int_0^T \epsilon dW(t) = \epsilon W(T)$. Since $P(\epsilon W(T)>K)>0$ I believe it MUST work but I found no way how to precisly formulate the "$\sim$" to get the result.
I have hoped that such simple statement must be somewhere but it looks like it is so trivial that noone cares to proof it. Closest thing I have found is Burkholder-Davis-Gundy inequality but I found no way to make use of it.
Thanks in advance for any hint.

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    $\begingroup$ Do you know anything else about $f$? For example, if $f \in L^2([0,T])$ then $X$ is normally distributed with mean 0 and variance $\int_0^T |f(t)|^2\,dt$, so your desired statement follows and you don't need the lower bound on $f$, except to exclude the trivial case where $f = 0$ a.e. If you don't have any control on the integrability of $f$, you may have trouble guaranteeing that the Ito integral exists. $\endgroup$ Jun 20, 2015 at 19:09
  • $\begingroup$ By the way, I wouldn't expect your $\sim$ statement to be true in any reasonable sense. The "infinitesimal increments" $dW(t)$ can be both positive and negative, so there's no reason why an inequality like $f > \epsilon$ should be preserved in any way when integrating. $\endgroup$ Jun 20, 2015 at 19:11
  • $\begingroup$ Thanks for your answer. $f$ is adapted process with some upper bound if needed. The reason for $f>\epsilon$ was that in fact I wanted to show that solution of $dX=f(X)dW$ may exceed any $K$. $\endgroup$ Jun 20, 2015 at 20:02
  • $\begingroup$ Oh, when you wrote $f(t)$ I assumed $f$ was deterministic. Of course if $f$ is a process then what I said does not apply. $\endgroup$ Jun 20, 2015 at 20:22
  • $\begingroup$ From my heuristic ideas I believe that $\sim$ may mean $P(X(T)>K) \geq P(\epsilon W(T)>K)$. But sill I have no way to prove it. $\endgroup$ Jun 21, 2015 at 8:31

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Welcome to MO.

You did not specify quantifiers in your question, so I take it to mean that given $T,\epsilon,K$ you look for a lower bound $\delta(T,\epsilon,K)>0$ on the probability in question.

Assume $|f(t)|\leq A$ for some deterministic $A$. You can represent the process $M_s=\int_0^s f(t) dW(t)$, $s\geq 0$, as a time change of a Brownian motion, that is $M_t=B_{\tau_t}$ for some Brownian motion $B$; Here $\tau_t=\int_0^t f^2(u) du$. (This is the Dambis, Dubins-Schwarz theorem). Now for a fixed $T$, $\tau_T\in [\epsilon^2 T, A^2 T]$ with probability 1. Thus, $$P(M_T>K)\geq P(B_u>K, \forall u\in [\epsilon^2 T, A^2 T])>0.$$

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