5
$\begingroup$

I want to forecast prices $S(t)$ of some asset based on historical daily values. I want to use the geometric Brownian motion given by an SDE: $$dS=\mu S t + \sigma S dB,$$ where $B$ is a Brownian motion, for modeling. The historical prices are $$\{S_i\}_{i=1}^N,$$ from which I calculate the log-returns ($N-1$ in total) $$Z_i=\ln\frac{S_i}{S_{i-1}},$$ and the 1-day historical volatility as the standard deviation of the returns: $$\hat{\sigma} = \sqrt{Var\{Z_i\}}.$$

Q1: Let's say I want to forecast the prices $S(t)$ for 180 days. Should I take $\sigma$ in the SDE as the 1-day volatility $\hat{\sigma}$ or as $\sqrt{180}\hat{\sigma}$? I'd say that $\sigma=\hat{\sigma}$ as I'm modeling day by day, but is it correct?

Q2: How do I compute the drift $\hat{\mu}$ from the historical prices? Is it simply $$\hat{\mu}=\frac{1}{N-1}\sum\limits_{i=1}^{N-1}Z_i,$$ and is it (the above formula) a 1-day drift?

$\endgroup$
  • $\begingroup$ You may want to take into account weekends and holidays if you are forecasting for half of a year. A trading year is about 252 days $\endgroup$ – user25064 Jul 20 '17 at 14:21
4
$\begingroup$

By looking at log returns, you are examing the stochastic process

$$ Q_t = \log S_t $$

given by

$$ \begin{align} dQ_t & = \left( \mu - \tfrac{1}{2}\sigma^2\right) dt + \sigma\, dB_t \\ & \equiv \alpha \,dt + \sigma\, dB_t \end{align} $$

where $\alpha=\mu-\tfrac{1}{2}\sigma^2$.

So far, everything is in continuous time. To interpret the SDE, you need to know how much time passes when $t$ increases by one unit. If the distance between $t=0$ and $t=1$ is one day, then $Q_{t+1}-Q_t$ is the daily log return, and $\mu$ is the daily drift. However, if the distance between $t=0$ and $t=1$ is one year, then $\mu$ is the annual drift.

Let's assume that one unit of $t$ is one day. Then defining $Z_i = Q_{i+1} - Q_i$ (which is equivalent to your definition of log returns) the formula

$$ \hat{\alpha} = \frac{1}{N-1} \sum_{i=1}^{N-1} Z_i $$

gives the one-day drift for this process, and

$$ \hat{\sigma}^2 = \frac{1}{N-2} \sum_{i=1}^{N-1} (Z_i - \hat{\alpha})^2 $$

gives the one-day variance (hence $\hat{\sigma}$ is the one-day standard deviation). To recover the estimator for the drift term $\mu$ you define

$$ \hat{\mu} = \hat{\alpha} + \tfrac{1}{2}\hat{\sigma}^2 $$

If you want the 180-day drift and standard deviation, you need

$$ \begin{align} \hat{\mu}_{180} & = 180\hat{\mu} \\ \hat{\sigma}_{180} & = \sqrt{180}\,\hat{\sigma} \end{align} $$

Purely as a point of notation, I would take your price observations to be $\{S_i\}_{i=0}^N$ so that you have $N+1$ price observations, and $N$ daily returns. It will simplify your formulas later.

$\endgroup$
  • 1
    $\begingroup$ What about the Ito component? shouldn't $\frac{\sum_{i=1}^{N-1}Z_i}{N-1}$ be an estimate of $\mu - \sigma^2 / 2$ $\endgroup$ – user25064 Jul 20 '17 at 14:17
  • $\begingroup$ Yes, you are completely correct - I'll update the answer. $\endgroup$ – Chris Taylor Jul 20 '17 at 14:25
  • $\begingroup$ So, I should input $\hat{\mu}_{180}$ and $\hat{\sigma}_{180}$ as estimates of $\mu$ and $\sigma$ in the SDE (eventually, I'll use the solution $S(t)$ to that equation); i.e., having an initial condition $S(0)$ I then just compute $S(180)$? $\endgroup$ – corey979 Jul 20 '17 at 14:29
  • $\begingroup$ @corey979 That depends on how you are interpreting $t$ in your SDE. See my updates to the answer. $\endgroup$ – Chris Taylor Jul 20 '17 at 14:36
  • $\begingroup$ Yes, $t$ is measured in days. I found in Mantegna & Stanley An Introduction to Econophysics, on p. 118, after Eq. 14.5, that $\mu$ and $\sigma^2$ are "per unit time", so the daily drift and volatility as the time unit is 1 day. And it makes more sense to make 2500\$ from the initial 1000\$ than an amount so huge that it causes an overflow on my computer. $\endgroup$ – corey979 Jul 20 '17 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.