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

Also, forgive me for cross-posting.

I have the following continuous time SDE for a 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 $0$ or $\mu$? How should I have defined the transformations? I suspect it might have something to do with the meaning of parameters and random variables. In the discrete time model above, $y_t$ is the mean-adjusted log return. In the SDE above that, $\mu$ probably means the interest rate plus half the variance (Questions on continuously compounded return vs long term expected return).

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    $\begingroup$ Hi Taylor. First notice that the limiting behaviour of discrete time processes and their continuous time counterpart is quite tricky (see the famous 1990 paper by Neslon). Second, the first SDE prevails under a certain probability measure which you have yet to define. This measure will give you the meaning of the parmeter $\mu$. Under $\Bbb{P}$ (real world measure) it's the expected return of the stock, under $\Bbb{Q}$ it's its funding rate (e.g. risk-free rate minus dividends minus repo rate). $\endgroup$ – Quantuple Jul 11 '18 at 7:45
  • $\begingroup$ Hi @Quantuple. Thanks for the note. Care to elaborate further in an answer? Also, which Nelson paper? There were two in 1990. $\endgroup$ – Taylor Jul 11 '18 at 16:55
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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

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  • $\begingroup$ However, your approach the probability of the stock price being negative is non-zero (though very small). One does not have such a problem when discretising the solution of the SDE. $\endgroup$ – FunnyBuzer Jan 13 at 12:11

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