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Let's decompose the return process $R_t$ as follows :

$$R_{t} = sign(R_{t}) * |R_{t}| $$

What's part of the equation is forecastable?

Richi Wa
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user1673806
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  • Just adding my opinion that this sounds like a homework question so you might not get your desired answer. I'd consider giving some intuition behind your question. – madilyn Mar 08 '13 at 10:29
  • Lol, i'm far from school from now even if i wish you were true about it. The reason of my question is follwoing this article http://jonathankinlay.com/index.php/2009/07/using-volatility-to-predict-market-direction/ that's arguing that sign is predictable while i ever heard that sign is not forecastable – user1673806 Mar 08 '13 at 10:34
  • Hi @user1673806 for every real number it holds that $x = sign(x) |x|$ so what is the model? What is the process? Or you ask differently: is the sign of a return (which market) forecastable (I would be rich if it was) or rather the magnitude of the return. The answer is: $|r_t|$ is usually auto-correlated and thus in a certain respect forecastable. This is what e.g. GARCH tries to do. Please improve the question. – Richi Wa Mar 08 '13 at 10:39
  • Maybe you include the link in the question. Anyways, I don't believe the first lines of it (I haven't read more). Would this mean if I know that markets go up today, I can forecast whether they go up or down tomorrow with satisfactory precision - certainly not in a tradable way. – Richi Wa Mar 08 '13 at 10:43
  • The article argue that volatility dynamics drive sign dynamics. Indeed, if Rt is normally distribued For a given mean return, m, the probability of a positive return is a function of conditional volatility. As the conditional volatility increases, the probability of a positive return falls – user1673806 Mar 08 '13 at 11:19
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    If you have a mean that it different from zero this might be true. But for daily periods - how different from zero will you mean be? do we talk about daily data? And if you "know" the mean then you already know a lot. by the way: rising volatility is often accompanied with falling prices (leverage effect) - this asymmetry is not captured. Bottom line: it is far more difficult in my mind. – Richi Wa Mar 08 '13 at 11:29
  • @user1673806, can you please correct your function in your question? You obviously meant to say R(t) not R(t+1) on the right hand side, correct? – Matt Wolf Mar 11 '13 at 03:48
  • I will correct that - it must be $t$ everywhere - otherwise it is confusing. – Richi Wa Mar 11 '13 at 08:29
  • Long range dependence in financial markets, by Rama Cont. By contrast, the autocorrelation function of absolute returns remains positive over lags of several weeks and decays slowly to zero. http://www.proba.jussieu.fr/pageperso/ramacont/papers/FE05.pdf, Long memory in return structures from developed markets, 2012, Sharad Nath Bhattacharya, MouSuMi Bhattacharya. http://www.ehu.es/cuadernosdegestion/documentos/110312sb.pdf – montyhall Mar 11 '13 at 16:22
  • @user1673806: If you liked one of the answers best you can accept it - Thank you :-) – vonjd Jul 12 '13 at 15:36

2 Answers2

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I think this one has a clear answer (I am solely talking about equities here):
The change magnitude is much more predictable than the direction.

The reason being that equity volatility is much more predictable than equity risk premiums. Volatility is nothing else but change magnitude and due to the stylized facts of volatility clustering together with mean reversion more predictable than the whole package, therefore also including direction. Basically the pattern is that there are phases when most movements are big in either direction and phases where everything is calm.

For a nice exposition see also this paper by Andrew Ang:
Equity market level

vonjd
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3

The two components you refer to in your questions are:

  • Market direction (the sign of the return)
  • Change magnitude (the absolute value of the return)

First, I'm sure you realize that neither of these are predictable at a 100%, otherwise there would be no way to make profit (you make profit by seeing things other didn't).

To answer the question, I would say that predicting the direction is a bit easier in a sense than the magnitude simply because of the possible outcomes:

  • the sign is a discrete variable with 2 possible outcomes;
  • whereas the magnitude is a continuous variable).

Other than that, there exists techniques for both, but neither will give you 100% accuracy, (or even close to that).

SRKX
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  • I am afraid I don't agree, please see my answer. What do you think? – vonjd Mar 11 '13 at 14:14
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    @vonjd Well I mean you can predict a range of magnitude, but a precise value is much more difficult because $P(|r_t|=x)=0 ~ \forall x$ in the continuous case. That's what I meant. But I see your point and I agree that volatility is more stable than market direction. – SRKX Mar 11 '13 at 15:16