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His model is not heteroskedastic. The log-error terms are i.i.d, so they have actually all the same variance. The distribution they are sampled from is a normal
by occamrazor 11y ago
His model is not heteroskedastic. The log-error terms are i.i.d, so they have actually all the same variance. The distribution they are sampled from is a normal variance mixture.
- QuantMash 11y agoInteresting. I'm no statistician but the blog post states that the authors of the original paper (Lo and Mckinlay) claim the test is heteroskedasticity-consistent. So what are you saying? Are you saying the original paper is wrong? That his example was wrong? (This seems more likely) And if the example is wrong, is it wrong to say it is heteroskedastic AND wrong to say it is a stochastic volatility model? Or just that it is heteroskedastic? As far as I can tell stochastic volatility simply means the variance itself is randomly distributed which looks consistent with his example? Just trying to clarift. Thanks.
- occamrazor 11y agoI haven't read the original paper, I can't comment on that. The model in the webpage however is not heteroskedastic (literally "unequal variance") because all the log-increments are iid. It could be legitimately considered a geometric Brownian motion with stochastic volatility, because the log-error is indeed normally distributed with variance picked from some stochastic distribution, in this case a normal distribution. This term however is normally used for models in which the volatility has more structure (e.g. the ARCH or GARCH models which are mentioned in the page).