0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\frac{autocorrelation + 1}{2}$

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.