I know I'm doing a short term forecast of a volatile time series using Monte Carlo, but I'm unsure as to the details - for example, I'm sure I had a very good reason for naming a term 'drift', but I can't recall why! I can't seem to find anything similar to what I'm doing when Googling for Monte Carlo forecast. Would anyone kindly help me by pointing me to some literature?

The pseudo-algorithm is:

Given highly volatile time series data $x_0, x_1, x_2, ..., x_n$

  1. Define $z_1, z_2, ..., z_n = ln(x_1/x_0), ln(x_2/x_1), ln(x_3/x_2),..., ln(x_n/x_{n-1})$.

  2. Define $\mu =$ average of $z$; $\sigma^2 = $ variance of $z$; drift $= \mu - \sigma^2/2$.

  3. Forecast using $x_{n+1} = x_n e^{d + \sigma R}$ where $d$ is the drift and $R$ is a random number generated from the inverse of the normal cdf with mean 0 and standard deviation 1.

  • 2
    $\begingroup$ You estimate the mean and variance of the volatility from the raw data and use the estimates in the prediction. I suggest you cut the first paragraph as it does not contribute to the question. $\endgroup$ – Xi'an Jan 18 '16 at 12:57
  • $\begingroup$ Thanks. Shortened paragraph as you suggested. I don't suppose you know where I can find a reference for the 'drift' term? $\endgroup$ – user101051 Jan 18 '16 at 13:16

If you consider a simple volatility model like$$x_{t}=x_{t-1}\exp\{z_t\}$$with $z_t\sim\mathcal{N}(\mu,\sigma^2)$, you get that $$\log(x_{t+1}/x_t)=z_{t+1}$$from which you can estimate $\mu$ and $\sigma$. Now, because $$\mathbb{E}[\exp\{z_t\}]=\exp\{\mu+\sigma^2/2\}$$ when $z_t\sim\mathcal{N}(\mu,\sigma^2)$ [reference] you get$$\mathbb{E}[\exp\{d+\sigma R\}]=\exp\{d+\sigma^2/2\}=\exp\{\mu-\sigma^2/2+\sigma^2/2\}=\exp\{\mu\}$$when $R\sim\mathcal{N}(0,1)$ which turns the forecast $$\hat{x}_{t+1}=x_t\exp\{d+\sigma R\}$$ into an autoregression in the sense that$$\mathbb{E}[x_{t+1}|x_t]=\exp\{\mu\}x_t$$

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  • $\begingroup$ Thanks so much for this explanation!! Can I ask, why is $\mathbb{E}[\exp\{z_t\}] = \exp\{\mu+\sigma^2/2\}$ and not $\mathbb{E}[\exp\{z_t\}] = \exp\{\mu\}$? Similarly, I'm unsure where the $\sigma R$ term on the lhs goes on the next line? Is it absorbed into the $+\sigma^2/2$ term? Thank you again! $\endgroup$ – user101051 Jan 18 '16 at 15:46
  • $\begingroup$ You have to solve the integral to understand why $\mathbb{E}[\exp\{z_t\}] = \exp\{\mu+\sigma^2/2\}$. And following your notations, $d+\sigma R\sim\mathcal{N}(d,\sigma^2)$. $\endgroup$ – Xi'an Jan 19 '16 at 9:29
  • $\begingroup$ Thank you very much for your patience and clear explanation Xi'an! $\endgroup$ – user101051 Jan 19 '16 at 19:53

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