5 ms·

It is quite a leap to go from 'not gaussian' to 'not random' as done here. All that has been falsified, as far as I can tell, is a very simple model of a random

by phreeza 11y ago

It is quite a leap to go from 'not gaussian' to 'not random' as done here. All that has been falsified, as far as I can tell, is a very simple model of a random walk with normally distributed disturbances. It would be interesting how much better it becomes if higher moments, in particular kurtosis ('fat tails') are included.

deleted 11y ago
[deleted]
- csours 11y ago
  https://news.ycombinator.com/newsguidelines.html https://news.ycombinator.com/newsguidelines.html > In Comments Be civil. Don't say things you wouldn't say in a face-to-face conversation. Avoid gratuitous negativity.
  - QuantMash 11y ago
    Apologies.
  - nborwankar 11y ago
    "It's free from interference. " "It suffers from interference by light."
QuantMash 11y ago
The article clearly states that the test extends to many forms of randomness beyond Gaussian: "Nevertheless, the desired effect of stochastic volatility namely, fatter tailed distributions ..." "... we want a test for the random walk hypothesis which passes (it concludes the market is random) even if the returns demonstrate heteroskedastic increments and large drifts. Why? Because both of these properties are widely observed in most historical asset price data (just ask Nassim Taleb) and neither invalidate the fundamental principle underpinning the random walk hypothesis, namely the Markov property (unforecastibility of future asset prices given past asset prices)"
- phreeza 11y ago
  Sorry, I missed that. Should have read it more closely! I got suckered by the mu, sigma and z-score terminology I suppose.
- 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 ago
    Interesting. 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 ago
      I 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).

by phreeza 11y ago

deleted 11y ago

[deleted]

csours 11y ago
https://news.ycombinator.com/newsguidelines.html https://news.ycombinator.com/newsguidelines.html > In Comments Be civil. Don't say things you wouldn't say in a face-to-face conversation. Avoid gratuitous negativity.
- QuantMash 11y ago
  Apologies.
- nborwankar 11y ago
  "It's free from interference. " "It suffers from interference by light."

QuantMash 11y ago

The article clearly states that the test extends to many forms of randomness beyond Gaussian: "Nevertheless, the desired effect of stochastic volatility namely, fatter tailed distributions ..." "... we want a test for the random walk hypothesis which passes (it concludes the market is random) even if the returns demonstrate heteroskedastic increments and large drifts. Why? Because both of these properties are widely observed in most historical asset price data (just ask Nassim Taleb) and neither invalidate the fundamental principle underpinning the random walk hypothesis, namely the Markov property (unforecastibility of future asset prices given past asset prices)"

phreeza 11y ago
Sorry, I missed that. Should have read it more closely! I got suckered by the mu, sigma and z-score terminology I suppose.
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 ago
  Interesting. 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 ago
    I 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).