Choosing a Return Mean for Physical-Measure GARCH Models

# Should stock prices be modeled as a martingale under the physical measure?


# Should stock prices be modeled as a martingale under the physical measure?












I need to simulate stock price under physical measure to assess the risk and find optimal parameters for put option insurance.

Simulating returns:

- Usually returns modelled as $r_t = \mu + \sigma_t z_t$ where $\mu$ is some constant, for example risk free rate. But there's a problem - it means $E[e^r_t] = e^{\mu + \sigma^2/2}$ constantly changing, depending on the volatility. The price process is not a martingale, that seems not very realistic.

- I think a proper way to model it would be $r_t = (\log E[e^r] - 0.5\sigma_t^2) + \sigma_t z_t$ where $E[e^r]$ could be something lke risk free rate.

For daily prices, the difference between two approaches is small, but for larger time periods, when time step is month, the $0.5\sigma_t^2$ term starts to be visible and results could be different.

GARCH fitting:

The related question - how to fit GARCH to historical prices? How to center error and predicted distribution.

- Constant location $\mu$:

$$ \epsilon_t = r_t - \mu\\ \text{likelihood} \sim N(loc=\mu, scale=\sigma_t) $$

- Constant expectation $E[e^r]$ (with $\sigma_{t-1}$ for error and $\sigma_t$ for likelihood).

$$ \epsilon_t = r_t - (\log E[e^r] - 0.5\sigma_{t-1}^2) \\ \text{likelihood} \sim N(loc=(\log E[e^r] - 0.5\sigma_{t}^2), scale=\sigma_t) $$

Question:

Can you please explain which approach is the correct one?

The difference may not be visible at daily time scale, but if GARCH used to predict longer horizons, 3m, 6m, 1y - the difference is visible.

Note: are physical prices martingale? - (in my opinion) while there's the risk premium, and so physical prices are not strictly martingale. Yet, they seems to be closer to martingale than not, and so modelling it as martingale is closer to reality and has smaller error. Or maybe there's a some compromise, say $E[e^r]-0.25\sigma^2$ (use 0.25 instead of 0.5)?

UPDATE: it seems GARCH-M address that issue by introducing adjustable risk premium $r_t = (\mu + \lambda g(\sigma_t)) + \sigma_t z_t$ where $g$ is some function of $\sigma$ possibly $\sigma^2$.

## Answer by QuantCalc.net (score 1)

https://quant.stackexchange.com/a/85291

Short answer is the first one is correct.

Here is the clean, practical way to fit a GARCH model to historical price data—from raw prices to estimated parameters.

- Convert Prices to Returns. To apply a GARCH model, historical prices must first be transformed into returns. We compute log-returns as $$ r_t = \ln(P_t / P_{t-1}), $$ which stabilizes variance and makes the series more suitable for volatility modeling. The main confusion might be from where. It doesn't make too much sense to keep $E[e^r]$ constant and prices don't not have to be martingale.

- Specify a GARCH(1,1) Model. A standard GARCH(1,1) assumes that returns follow $$ r_t = \mu + \epsilon_t$$with conditionally normal innovations $$ \epsilon_t \mid F_{t-1} \sim N(0, h_t) .$$The conditional variance evolves through $$ h_t = \omega + \alpha \epsilon_{t-1}^2 + \beta h_{t-1}, $$ capturing volatility clustering.

- Log-Likelihood Function. Given the conditional normal assumption, the log-likelihood of observing the data is obtained by summing the log of the corresponding Gaussian densities: $$ \ell(\theta) = \sum_{t=1}^T \left[ -\tfrac12 \big( \ln(2\pi) + \ln h_t + \epsilon_t^2 / h_t \big) \right]. $$ This is the objective function maximized during parameter estimation.