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I have a Bernoulli process $\Phi(t)$ with a symmetric distribution $p=1/2$. The random variable can take values $a,b$. My question is what is the covariance of this process $\langle\Phi(t)\Phi(t')\rangle$?


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@Sasha: No, it isn't. I am computing a more complex stochastic process $dX(t)=dW(t)\Phi(t)$. – Jon Jan 18 '12 at 14:47
There is not enough information to determine the covariance (or autocovariance function) of the process. $\text{Cov}(\Phi(t), \Phi(t'))$ can take on various values in $[-(b-a)^2/4, (b-a)^2/4]$ depending on the (unspecified) relationship between the random variables $\Phi(t)$ and $\Phi(t')$. – Dilip Sarwate Jan 18 '12 at 14:55
@DilipSarwate: Fine. You can consider this the flipping of the sign in a Brownian motion. – Jon Jan 18 '12 at 15:03
@Jon Please see updates to both this and the previous question of yours. – Sasha Jan 18 '12 at 17:15
up vote 2 down vote accepted

Per your comment, let $\Phi(t) = \operatorname{sign}(W(t))$. Using results from your previous question it would be $$\begin{eqnarray} \mathbb{E}(\Phi(t) \Phi(t^\prime)) &=& \mathbb{P}(W(t) \geqslant 0, W(t^\prime)\geqslant 0) + \mathbb{P}(W(t) < 0, W(t^\prime) < 0) \\ &\phantom{=} &- \mathbb{P}(W(t) < 0, W(t^\prime)\geqslant 0) - \mathbb{P}(W(t) \geqslant 0, W(t^\prime) < 0) \\ &=& \frac{2}{\pi} \arcsin\left( \sqrt{\frac{\min(t,t^\prime)}{\max(t,t^\prime)}} \right) \end{eqnarray} $$

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Thank you very much! – Jon Jan 18 '12 at 15:24
For example, when $t=t'$ the RHS is $\frac34$... Sorry but it seems this (accepted) answer needs some revision (as well as the (accepted) answer on the other page). – Did Jan 18 '12 at 16:59
@DidierPiau Thank for catching these embarrassing mistakes. I have updated both posts with correct expressions. – Sasha Jan 18 '12 at 17:14
@Sasha: This does not seem really intuitive. I mean if $t\gg t'$ this goes to 0 while for Brownian motion is seen increasing like $t$. Besides, I would expect that in this same limit the covariance should go rather like the variance of the process. – Jon Jan 18 '12 at 20:13
@Jon Conditional of $W(t^\prime)$, $W(t) \sim W(t^\prime) + W(t-t^\prime)$. As $t - t^\prime$ is the scale of fluctuations, then $t-t^\prime$ becomes much bigger than $W^(t^\prime)$, the disparity of probabilities $\mathbb{P}(W(t) \geqslant 0 | W(t^\prime))$ and $\mathbb{P}(W(t) < 0 | W(t^\prime))$ dwindles. So it seems intuitive to me. – Sasha Jan 18 '12 at 20:27

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