For Simplicity $ (\Omega ,\mathbb F,\mathbb P)$ is a probability space,
the S.D.E is $dX_t=b(X_t)dt+dB_t$ and it's 1 dimensional,
$b(X_t)$ is Lipschitz Continuous (and let's assume with constant 1), also assume $X_0 =x$.
Can one help me prove that if $X^x_t$ is a solution, then $P(X_t\geq M)>0 $ for all positive M,t.
My attempt:
By $b$ being lipschitz we have that the solution is strong and unique. Assume by contradiction that: $\mathbb P (X_t^x>M)=0 \rightarrow \mathbb P(X_t^x\leq M)=1$
Which means that $\int_0^M X_t^xd\mathbb P=1$ which I'm sure is almost a contradiction, but to what exactly?
There are many ways to show this. Perhaps, this is the simplest one: there is a constant $C$ such that $b(x) > -C(x+1)$ for all $x\in \mathbb R$. Consider the equation $$ dY_t = -C(Y_t +1) + dB_t\tag{1} $$ with the initial condition $Y_0 = y<x$. It is not hard to show that $Y_t <X_t^x$ for all $t\ge 0$. But equation (1) is easy to solve: $$ Y_t = e^{-Ct}\left(y+1 + \int_0^t e^{Cs}dB_s\right)-1. $$ In particular, for any $t\ge 0$, $Y_t$ has a Gaussian distribution. Therefore, it is not bounded from above, so neither is $X_t$. The unboundedness from below is shown similarly.