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 $z_t\sim\mathcal{N}(\mu,\sigma^2)$ [\[reference\]][1] 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$$\mathbb{E}[x_{t+1}|x_t]=\exp\{\mu\}x_t$$


  [1]: https://en.wikipedia.org/wiki/Log-normal_distribution