Mapping Binary-Series Autocorrelation to Repeat Probability
Summary
The document derives how lag-one autocorrelation relates to the chance that a binary series repeats its current value. It assumes observations are either +1 or −1, each occurs with probability 0.5, and the transition process is symmetric: the probability of moving from +1 to +1 is the same as moving from −1 to −1. Under these assumptions, the conditional expectation of the next observation is expressed in terms of the repeat probability, then related to the product of adjacent observations.
The resulting relationship is that the repeat probability equals one half of one plus the autocorrelation. This recovers the stated boundary cases: autocorrelation of −1 corresponds to no repeats, zero corresponds to a 50% repeat chance, and +1 corresponds to certain repetition. The mapping depends on the balanced, symmetric binary setup; autocorrelation alone does not generally determine transition probabilities for other distributions or asymmetric transition patterns.
Key ideas
- For a balanced binary series with symmetric transitions, lag-one autocorrelation is twice the repeat probability minus one.
- The probability of repeating the current sign is one half of one plus the autocorrelation.
- An autocorrelation of zero implies a 50% repeat probability under the stated assumptions.
- The result relies on equal marginal probabilities for +1 and −1 and symmetric transition probabilities.
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Full text
# Interpreting Autocorrelation as probability
# Interpreting Autocorrelation as probability
I was recently asked:
Given a random time series of 1s and -1s. Eg of a sample = [1, 1, 1, -1, -1, 1, -1,..]. The autocorrelation of this series is Z. What can you say about the probability of a 1(or -1) followed by 1 (or -1 respectively)?
We can further assume that probability of +1 and -1 is 0.5 respectively.
One thing is for clear, if Z is -1, the probability of 1(-1) followed by 1(-1) is 0 and if Z is 1, the probability is 1. Can we somehow use Z to determine the probability of repeated occurrence as the question asked?
Thank you.
[Note]: The basis of my questions comes from the following observation. If the autocorrelation is -1, then the probability of successive outcome is 0, if the autocorrelation is 0, the probability of successive outcome is 0.5 and if the autocorrelation is 1, the probability of success outcome is 1. I was wondering if this mapping from autocorrelation to probability can be interpolated between the key points above.
## Answer by vpy (score 0, accepted)
https://quant.stackexchange.com/a/53360
Assume $P(X_{t+1}=1| X_t=1) = q ; P(X_{t+1}=-1| X_t=1) = 1-q $
Then, $E[X_{t+1} | X_t = 1] = 1*q + (-1)(1-q) = 2q-1$
$E[X_t,X_{t+1}] \\ = E[E[X_t,X_{t+1} | X_t]] \\ = E[X_t,X_{t+1} | X_t = 1]*P(X_t = 1) + E[X_t,X_{t+1} | X_t = -1]*P(X_t = -1) \\ = 1*E[X_{t+1} | X_t = 1]*0.5 + (-1)*E[X_{t+1} | X_t = -1]*0.5 \\ = 0.5 * ( E[X_{t+1} | X_t = 1] - E[X_{t+1} | X_t = -1] ) \\ = 0.5 * (2q-1 - (1-2q)) = 2q-1 $
Hence, probability of successive occurrence (q in the proof above) is equal to $\frac{autocorrelation + 1}{2}$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.