Let $b: \mathbb{R} \rightarrow \mathbb{R}$ be a Lipschitz-continuous function and let $X_t$ be a real valued stochastic process satisfying the stochastic differential equation $dX_t= b(X_t) dt+ dB_t$, $X_0=x$. Prove that for any $M> 0$, $t> 0$ and $x \in \mathbb{R}$ we have that $P(X_t \geq M)>0$ but in the case that $b(x)= \alpha$ for some $\alpha <0$ we have that $P(\lim_{t \rightarrow \infty} X_t= - \infty)=1$.
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1What do you know? What did you try? Where are you stuck? (Sounds like homework, is it homework?) – Did Jan 08 '12 at 22:32
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I believe the first part follows from Girsanov's theorem. For the second part, you can solve the SDE explicitly, and then the law of iterated logrithms works. – Aaron Jan 09 '12 at 09:37
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As mentioned in Prop 3.10 in Shreve-Karatzas we have that for the martingale
$$Z_{t}:=exp\left(-\int_{0}^t b(X_{s})dW_s-\frac{1}{2}\int_{0}^t b^{2}(X_{s})ds \right)$$
and measure $\frac{dQ}{dP}=Z_{T}$, the process
$$X_{t}=X_{0}+\int_{0}^t b(X_{s})ds+W_{s}$$
is a Brownian motion $\tilde{W}_{t}$. So
$$E\left(1_{X_{t}\geq M}Z_{T}\right)=Q(\tilde{W}_{t}\geq M)>0,$$
which implies $P[X_{t}\geq M]>0$.
For the particular case of constant negative drift, we have the exact solution
$$X_{t}=x-|\alpha| t+B_{t}.$$
and so here we use that $\lim_{t\to +\infty}\frac{B_{t}}{t}=0$ to in fact get
$$\lim_{t\to +\infty}\frac{X_{t}}{t}=-|\alpha|.$$
Thomas Kojar
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