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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 $$\mathbb{E}[\exp\{z_t\}]=\exp\{\mu+\sigma^2/2\}$$ you get$$\mathbb{E}[\exp\{d+\sigma R\}]=\exp\{\mu-\sigma^2/2+\sigma^2/2\}=\exp\{\mu\}$$which turns the forecast $$\hat{x}_{t+1}=x_t\exp\{d+\sigma R\}$$ into an autoregression in the sense that$$\mathbb{E}[x_{t+1}|x_t]=\exp\{\mu\}x_t$$