I am struggling to understand the intuition and use of half-life period of a mean reverting process. According to definition, half-life period shows how long it takes for a time series to return halfway to its mean after a deviation. Below I summarize my questions:

  1. Why half-life? Why we are interested to know how long it takes for a time series to return halfway to its mean after deviation, but not to its mean? Isn't it more important knowing how long it takes to revert to the mean?
  2. According to definition, $HL=-\frac{ln(2)}{\theta}$, where $\theta$ is the speed of mean reversion of an OU process. Half-life does not depend on the current value of the process and it is constant. Does this make sense? Suppose mean of the mean reverting process is $0$, and the estimated $HL$ is 10 days. Then, if current value of the process is 2,000 then we say it will go to 1,000 within 10 days, and from 1,000 to 500 again within 10 days? Shouldn't from 2,000 to 1,000 take longer than from 1,000 to 500?
  3. How useful is $HL$? How it can be applied to make trading decisions?

2 Answers 2

  1. In processes such as OU, it takes infinite time to revert to the mean completely. An unperturbed process starting at some point away from the mean asymptotes towards the mean without ever touching it. Thus, specifying the time to revert to the mean ($\infty$) is not informative, as that does not discriminate between different processes that share that feature.

  2. Indeed, the speed of reversion is inversely proportional to the distance between the current point and the mean. If you are far away, you will be reverting quickly, but you will be slowing down as you approach the mean. Thus it does not take longer to reduce the gap from 2000 to 1000 than from 1000 to 500. This is how OU-type processes work.

  3. HL allows you to compare different processes. If you want to make money by betting on a process reverting to its mean, there are quick ways (when HL is short) and slow ways (when HL is long). If you have to wait twice as long by betting on process 2 than on process 1, your expected log-return per unit of time is accordingly twice lower.

  • $\begingroup$ Thanks a lot, very helpful. One more remark. Let's assume current value is 1,000 (mean is still 0) and yet we observe upward trend. In other words, the process is diverging more from the mean, rather than gravitating towards mean. Still we have 10 days until it comes to 500? HL does not distinguish whether it is coming back to the mean or diverging more from the mean, right? $\endgroup$
    – Sane
    Commented Jan 10 at 9:01
  • $\begingroup$ @Sane, HL is what happens on average. In concrete trajectories the realized HL will be higher or lower. Also, if you have an estimated HL of 10 and then observe what happens over the next 3 days, you should update your expectation of when half of the gap will be closed with the new information (the 3 days). $\endgroup$ Commented Jan 10 at 10:37
  • $\begingroup$ I see, thank you. $\endgroup$
    – Sane
    Commented Jan 10 at 10:38
  • 1
    $\begingroup$ @Sane - two articles you may find useful: Bertram, William K. "Optimal trading strategies for Itô diffusion processes." Physica A: Statistical Mechanics and its Applications 388.14 (2009): 2865-2873; and, Bertram, William K. "Analytic solutions for optimal statistical arbitrage trading." Physica A: Statistical mechanics and its applications 389.11 (2010): 2234-2243. $\endgroup$
    – krkeane
    Commented Jan 10 at 15:48
  • $\begingroup$ @krkeane Thanks a lot, will have a look! $\endgroup$
    – Sane
    Commented Jan 12 at 7:30

It might be helpful to consider an equivalent but different definition of the half life.

Suppose we have a standard (zero-mean) OU process defined by $dx_t = -\theta x_t dt + \sigma dW_t$ (which is how Wikipedia does it).

We could define a half-life at time $s$ as the time $h$ such that $𝔼(x_h|x_s)=\frac{1}{2}x_s$. That is, how long until half the deviation from the mean is, in expectation, lost.

This is easy to compute because \begin{align} \frac{1}{2}x_s & = 𝔼(x_h|x_s) \\ & = x_s\exp(-\theta h) \end{align} and we can cancel the $x_s$ terms to give \begin{align} \frac{1}{2} = \exp(-\theta h) \end{align} which rearranges to \begin{align} h = \frac{\ln(2)}{\theta} \end{align} The cancellation of $x_s$ means that a half-life does not depend on either $s$ or $x_s$ and so we are entitled to call it "the" half-life of the process. Your definition of half-life has a negative sign but I suspect that your associated OU process is probably defined slightly differently.

So, to your point 2, the half-life is a constant for an OU process. Of course, for any actual observed process you have a noise term that means that the smooth reversion of the expectation is not realized and you may achieve or overshoot the mean in finite time. In fact, I'm sure that there are some results giving the distribution of the time until the mean is next hit.

To your point 1, the main use of the half-life is to restate the $\theta$ parameter into more intuitive terms. Certainly, when presenting this stuff to non-technical audiences, the concept of half-life seems a lot easier for people to understand than the rather abstract $\theta$.


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