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$$ dz_t \sim O\left(\sqrt{dt}\,\right) $$

$z$ is a Brownian motion random variable, for reference. I just don't understand what the $\sim O$ part means. I've looked up the page for Big O notation on wikipedia because I thought it might be related, but I can't see the link.

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    $\begingroup$ This notation should be shot on sight (independently of the context, already $\sim$ and $O$ together is a crime...). What is your source? $\endgroup$ – Did Oct 11 '13 at 16:45
  • $\begingroup$ I'm talking an online course on asset pricing (on coursera, taught by John Cochrane) and this is on the first page of the first set of notes. This line in context is here: i.imgur.com/qWhlIBl.png $\endgroup$ – Nish Oct 11 '13 at 17:03
  • $\begingroup$ This question may be helpful: $dB_t^2 = dt$ $\endgroup$ – saz Oct 12 '13 at 10:55
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This notation is for intuition about the change in brownian motion. That is, this is not formal but is a good enough mnemonic for practitioners.

$dz_t$ is the change in brownian motion at time $t$. Lets informally think of this as $z_{t+dt} - z_t$. We know by the properties of brownian motion that this is normally distributed with variance $t+dt - t = dt$. Thus, informally again, $dz_t$ is zero-mean normal with standard deviation $\sqrt{dt}$.

That is, $dz_t$ "grows" as $\sqrt{dt}$. Loosely speaking again, for a zero mean random variable with standard deviation $c$, a lot of the probability mass is within the interval $[-c,c]$. This is captured by the Big-O notation.

In conclusion, the mnemonic is giving the intuition that if you see the $dz_t$ term (called the diffusion term) while describing a stochastic process in ito calculus, you should think of that term as being $O(\sqrt{dt})$ compared to the drift term which is O($dt$).

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  • $\begingroup$ Reasoning based on the fact that $E[|z_{t+s}-z_t|] = c\sqrt{s}$ may be a bit confusing because of the use of absolute value within the expectation as posted by @Did. $\endgroup$ – Theja Oct 11 '13 at 17:50
  • $\begingroup$ Sorry? I do not get your comment... $\endgroup$ – Did Oct 11 '13 at 18:56
  • $\begingroup$ My reservation was about the use of $|z_{t+s} - z_t|$ vs $z_{t+s} - z_t$ in the explanation. Maybe it is intuitive that way. Nonetheless, your explanation is precise and to the point. $\endgroup$ – Theja Oct 11 '13 at 19:15
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My advice about this notation is already in a comment... What the notation refers to is the fact that if $(z_t)$ is a linear Brownian motion then, for every $t$, $z_{t+s}-z_t$ is normal centered with variance $s$ hence, for example $E[|z_{t+s}-z_t|]=c\sqrt{s}$ for some absolute constant $c$.

Note that pathwise properties of $(z_t)$ are quite different since, for example, almost surely, $$ \liminf_{s\to0+}\frac{z_{t+s}-z_t}{\sqrt{2s\log\log1/s}}=-1,\qquad\limsup_{s\to0+}\frac{z_{t+s}-z_t}{\sqrt{2s\log\log1/s}}=1. $$

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