The time series is governed by the equation $S(T)=S(0)e^{(\mu-\frac{\delta^2}{2})T+\delta(w(T)-w(0))}$, in which $w(t)$ is a standard Brownian motion. Now given the data $\{S(t)\}_{t=0}^{t=T}$, how to estimate $\delta$ and $\mu$?

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    $\begingroup$ $(\mu-\frac{\delta^2}{2})t$ correct? (I added the $t$) $\endgroup$ – Cam.Davidson.Pilon Jun 13 '13 at 17:52

By considering the log of the time series, i.e.

$$ \{\log{S(t)}\}_{t=0}^{t=T}$$

we have

$$ \log{S(0)} + (\mu - \delta^2/2)t + \delta w(t) $$

( $w(0) = 0$ ). Taking first differences of this series, $\log{S_{t_i}} - \log{S_{t_{i-1}} }$, gives a new series:

$$ (\mu - \delta^2/2)( t_i - t_{i-1} ) + \delta( w(t_i) - w(t_{i-1} ) )$$

This new series is independent and Normally distributed, with mean $(\mu - \delta^2/2)( t_i - t_{i-1} )$ and standard deviation $\delta\sqrt{ t_i - t_{i-1} }$. One can use MLE to find the "best" estimators of the two unknowns.

  • $\begingroup$ Thanks! But can you be more specific on how to use MLE to estimate those two unknowns? $\endgroup$ – Ziqian Xie Jun 13 '13 at 18:05
  • $\begingroup$ I should ask before I begin: is this homework/assignment? Can you show me your thoughts? $\endgroup$ – Cam.Davidson.Pilon Jun 13 '13 at 18:06
  • $\begingroup$ Thx, I know how to do it now. $\endgroup$ – Ziqian Xie Jun 13 '13 at 18:17
  • $\begingroup$ Urukec, if that's for work related to some course, please tag it self-study, even if you're done. $\endgroup$ – Glen_b Jun 13 '13 at 22:58

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