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GARCH(1,1)

In discrete time, we can model returns as follows \begin{align} r_t &= \mu + \sigma_t\epsilon_t\\ \sigma_t^2 &= \omega + \alpha \epsilon_{t-1}^2 + \beta\sigma_{t-1}^2 \end{align}

  • Returns are driven by white noise ($\epsilon_t\overset{iid}{\sim} N(0,1)$)
  • The GARCH process converges to $d\sigma_t^2=\kappa(\theta-\sigma_t^2)dt+\xi\sigma_t^2dW$ (Nelson, 1990)

Heston

In continuous time, we can model returns as follows \begin{align} \frac{dS}{S} &= \mu dt+\sigma_tdW_t \\ \sigma_t^2 &= \kappa(\theta-\sigma_t^2)dt+\xi\sigma_tdW_t^2 \end{align}

  • Returns are driven by white noise ($dW_t\sim N(0,dt)$)
  • This is pretty much the limit of the GARCH process with the exception of $\sigma_t$ instead of $\sigma_t^2$ in the $dW_t^2$ part

Skewness and Kurtosis

  • In the discrete GARCH model, higher moments of returns are non-trivial, see this answer by @RichardHardy
  • In the continuous model, higher moments of returns are zero: $\mathbb{E}\left(\left(\frac{dS}{S}\right)^3\right)=0$. Essentially, the Ito process $dS$ has finite quadratic variation and therefore all higher moments are zero.

Question: Why does the discrete time volatility model allow us to realistically capture kurtosis whereas the time continuous model does not?

In quant finance, discrete time and continuous time models often fit well (e.g. random walk $\to$ Brownian motion, binomial tree $\to$ Black-Scholes). Why do these models differ? I understand that a GARCH model has predictable volatility whereas the Heston model has true stochastic volatility, but I'm not sure that's the explanation for the zero kurtosis in the time continuous model?

Note: the same ideas (finite quadratic variation $\Rightarrow$ zero skewness and kurtosis) apply if we use GARCH to model log-returns and consider $d\ln(S)$ instead of $\frac{dS}{S}$

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  • $\begingroup$ [Sorry for editing the question rather than commenting] The moment-generating function in the Heston model for $\ln(S)$ is known in closed-form. So you can compute its derivatives and evaluate it zero to get the moments of $\ln(S)$. That would reveal a non-zero skewness and excess kurtosis. But that does not seem to relate directly to the percentage returns you are interested in, $\frac{\text{d}S}{S}$ $\endgroup$ – Kevin May 5 at 23:29
  • $\begingroup$ Wouldn‘t you need to equate the time horizon between both processes to be able to make comparisons? Ie what is the skewness induced b Heston for one trading day vs ehh does the discrete model say for one step (which equates to one day here). $\endgroup$ – Kermittfrog May 6 at 4:47
  • $\begingroup$ @Kermittfrog yes, there’s a difference between discrete time and continuous time but I don’t see why intuitively discrete time models are so much better (allow for skewness and kurtosis). I would have guessed that both models give similar results. Isn’t the Heston model inadequate because it suggests stock returns are symmetric and have no fat tails? $\endgroup$ – Alex May 6 at 7:26
  • $\begingroup$ @Kermittfrog You’re suggesting that if I use a Euler discretization, then I may get non-zero skewness and kurtosis for percentage returns. But the question is why do these moments vanish if $\Delta t\to0$? Why do time continuous models fail to account for skewness and fait tails? $\endgroup$ – Alex May 6 at 7:48
  • $\begingroup$ Think about it the other way, then: the continuous time limit of the GARCH process looks non-skewed as well, doesn't it? Hence I think that you need some 'time' to accumulate skewness in such a model. $\endgroup$ – Kermittfrog May 6 at 9:30

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