Monte Carlo rolling forecast of time series - details needed

Question

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.

Answer 1

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$$\mathbb{E}[x_{t+1}|x_t]=\exp\{\mu\}x_t$$