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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?

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    $\begingroup$ 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 $n\geqslant1$. $\endgroup$ – Did Jan 4 '16 at 15:59
  • $\begingroup$ @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? $\endgroup$ – RTJ Jan 4 '16 at 18:08
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    $\begingroup$ 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 $N$"). $\endgroup$ – Did Jan 4 '16 at 18:10

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