Hot answers tagged


In this context, unconditional variance refers to the stationary variance level predicted by your GARCH model. This quantity need not coincide with the sample variance of the data on which the latter model has been calibrated. That being said, in an effort to reduce the complexity of the GARCH parameters' estimation process (nasty non-linear optimisation ...


it certainly works best at the money. Why? I think it comes from the fact that Black's formula is approximately linear at the money. The approximation $$ \frac{1}{\sqrt{2\pi}} \operatorname{SR} \sigma \sqrt{T} A, $$ with $A$ the annuity is remarkably good. One way of deducing these formulas is to do an asymptotic/Taylor expansion about $\sigma=0.$


Let $X^h$ be your hourly process Let $X^d$ be your daily process Let $\delta$ be one day you have $$X^d_t=\frac{1}{\delta}\int_{t-\delta}^{t}X^h_s ds$$ $$dX^h_t = a(b-X^h_t)dt + \sigma dB_t$$ $$\Delta X^d_t := X^d_{t+\delta}-X^d_t =\frac{1}{\delta}\int_{t-\delta}^t\left(X^h_{u+\delta}-X^h_{u}\right)du$$ so it is a gaussian random variable by knowns ...


You can aggregate your starting hourly data to obtain daily data and re-estimate the parameters, then simulate. Alternatvely, with your parameters already obtained, you can simulate hourly data and make a post-simulation aggregation to have daily data.


You are on the right track IMO, except that there are no exceptions to "the greater the expected value, the higher the option value" rule, since an option premium's is precisely the discounted expectation of its payout by absence of arbitrage - at least under the risk-neutral measure. As far as the influence of volatility on the option value is concerned: ...

Only top voted, non community-wiki answers of a minimum length are eligible