# Discretizing a Stochastic Volatility SDE

How does the discrete time stochastic volatility model arise from the continuous time one?

I have the following continuous time stochastic volatility model. $$S_t$$ is the price, and $$v_t$$ is a variance process. $$dS_t = \mu S_tdt + \sqrt{v_t}S_t dB_{1t} \\ dv_t = (\theta - \alpha \log v_t)v_tdt + \sigma v_t dB_{2t} .$$ I'm more familiar with the discrete time version: $$y_t = \exp(h_t/2)\epsilon_t \\ h_{t+1} = \mu + \phi(h_t - \mu) + \sigma_t \eta_t \\ h_1 \sim N\left(\mu, \frac{\sigma^2}{1-\phi^2}\right).$$ $$\{y_t\}$$ are the log returns, and $$\{h_t\}$$ are the "log-volatilites." Keep in mind there might be some confusion about parameters; for example the $$\mu$$s in each of these models are different.

How do I verify that the first discretizes into the second?

Here's my work so far. First I define $$Y_t = \log S_t$$ and $$h_t = \log v_t$$. Then I use Ito's lemma to get \begin{align*} dY_t &= \left(\mu - \frac{\exp h_t}{2}\right)dt + \exp[h_t/2] dB_{1t}\\ dh_t &= \left(\theta - \alpha\log v_t - \sigma^2/2\right)dt + \sigma dB_{2,t}\\ &= \alpha\left(\tilde{\mu} - h_t \right)dt + \sigma dB_{2t}. \end{align*}

I got the state/log-vol process piece. I use the Euler method to discretize, setting $$\Delta t = 1$$, to get \begin{align*} h_{t+1} &= \alpha \tilde{\mu} + h_t(1-\alpha) + \sigma \eta_t \\ &= \tilde{\mu}(1 - \phi) + \phi h_t + \sigma \eta_t \\ &= \tilde{\mu} + \phi(h_t - \tilde{\mu}) + \sigma \eta_t. \end{align*}

The observation equation is a little bit more difficult, however:

\begin{align*} y_{t+1} = Y_{t+1} - Y_t &= (\mu - \frac{v_t}{2}) + \sqrt{v_t}\epsilon_{t+1} \\ &= \left(\mu - \frac{\exp h_t}{2} \right) + \exp[ \log \sqrt{v_t}] \epsilon_{t+1} \\ &= \left(\mu - \frac{\exp h_t}{2}\right) + \exp\left[ \frac{h_t}{2}\right] \epsilon_{t+1}. \end{align*} Why is the mean return not $$0$$?

• Be aware: crossposted. Jul 11, 2018 at 15:04

I guess you can discretize the raw price process too instead of the log price process. You get $$S_{t+1} = S_t + \mu S_t + \sqrt{v_t} S_t Z_t$$ (where $$Z_t$$ is a standard normal variate), or $$\frac{S_{t+1}}{S_t} - 1 = \mu + \sqrt{v_t} Z_t.$$ Got the idea from: https://arxiv.org/pdf/1707.00899.pdf