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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 $P(\lim_{t \rightarrow \infty} X_t= - \infty)=1$.

nick
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    What do you know? What did you try? Where are you stuck? (Sounds like homework, is it homework?) – Did Jan 08 '12 at 22:32
  • 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

1 Answers1

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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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