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I was reading Oksendal's book "Stochastic Differential Equations", fifth Ed., pp. 62-63 and came across some counter-intuitive properties of the Geometric Brownian motion (GBM).

Let $\alpha,r>0$ and $B_t$ a Brownian motion. Then, the GBM $$N_t=N_0 \exp((r-\alpha^2/2) t+\alpha B_t)$$ is the solution of the SDE $$dN_t=rN_tdt+\alpha N_tdB_t$$
It is proved (page 63) that $$E[N_t]=E[N_0] e^{rt}$$ which means that the mean of the process is exponentially increasing over time for $r>0$. On the other hand, if $r<\alpha^2/2$ it is proved that $$\lim_{t\rightarrow\infty}N_t=0 \qquad a.s.$$

These two properties seem intuitively contradictory (the mean increases exponentially over time but at the same time the process converges to zero a.s.). I was wandering if someone can provide some physical explanation on this.

Also, a follow up question: For $0<r<\alpha^2/2$ is the solution bounded a.s. or bounded in the mean?

RTJ
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    For a sequence of positive random variables such that $X_n\to0$ almost surely and yet, $E(X_n)\to\infty$, try $P(X_n=n^4)=1/n^2$, $P(X_n=1/n^2)=1-1/n^2$ for every $n\geqslant1$. – Did Jan 04 '16 at 15:59
  • @Did Thank you. So this actually means that as $t$ grows to infinity the probability of $N_t$ taking large values goes to zero with a slower rate. What does this imply for the actual realization of the process? What will we observe in simulations? – RTJ Jan 04 '16 at 18:08
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    In the example in my comment, after a while, each $X_n$ equals $1/n^2$ (but beware that "after a while" means here, "for every $n\geqslant N$ for some random, almost surely finite $N$"). – Did Jan 04 '16 at 18:10

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I think the limit

$$\lim_{t\rightarrow\infty}N_t=0 \qquad a.s.$$

is clear since by iterated law of the logarithm the Brownian motion grows much closer than linear ($\sqrt{t\log\log t}$) and so it just says that the negative linear drift beats it.

In the computation

$$N_t=N_0 \exp((r-\alpha^2/2) t+\alpha B_t)$$

we use that $B_{t}\sim N(0,t)$. But as you know with Gaussians, most of the mass is concentrated near the mean and it reaches $\pm t$ with increasingly small probability. So that is why most of the actual paths that you simulate will behave according to the zero-limit, the same that sampling from a Gaussian mostly gives points centered around zero.

Also, a follow up question: For $0<r<\alpha^2/2$ is the solution bounded a.s. or bounded in the mean?

The mean is unbounded as you just computed. The solution is not necessarily bounded for finite time. The limit just says that eventually it goes to zero after some random time.

Even in the case of $r=0$ from here Density of first hitting time of Brownian motion with drift we have that the density of hitting any level $a>0$ is positive. See here too Geometric Brownian Motion Probability of hitting uper boundary.

Thomas Kojar
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