Let's say we have geometric Brownian motion:
$$
dS_t = \mu S_tdt + \sigma S_tdW_t
$$
Then the SDE becomes:
$$
S_t = S_0\exp\left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t\right)
$$
Say $\mu$ is zero and the drift is zero. But below, the drift term $(\mu - \sigma^2)t$ becomes $(-(\sigma^2)/2)t$, which will make a drift occur. What gives? Wouldn't it make more sense to drop the drift component entirely?
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$\begingroup$ More sense in terms of modelling some specific real-life phenomenon? Or is the question about why the solution to $dS_t = \sigma\,S_t\,dW_t$ is not $S_t = S_0\, \exp (\sigma\,W_t)$? $\endgroup$– Juho KokkalaOct 2, 2015 at 6:31
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$\begingroup$ The latter. Just looking at the formula that is derived, it implies a drift over time. If your drift is zero to begin with, why would there be a residual drift in the derived formula? $\endgroup$– TH4454Oct 8, 2015 at 21:42
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2 Answers
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Your logic is flawed, as a driftless SDE would still have quadratic variation applied to it. The quadratic variation would add a term $\sigma$, while $1/2$ comes from ito Lemma.
A martingale under a probability measure doesn't mean the solution wouldn't have a "drift" term.
answered Jul 19, 2018 at 4:53
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You need to be a bit more careful about the fact this is geometric, not additive, Brownian motion.
Were it additive the solution would be a family of normally distributed RVs as a function of time.
But GBM solves to give a log-normal distribution. And the problem, in a nutshell, is that the expected value of a log-normal distribution is not simply $\mu$; it's median turns out to be $\mu$. See: https://en.wikipedia.org/wiki/Log-normal_distribution
