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I am trying to get an understanding of autocorrelation and I am having some issues with trying to understand the process.

I have a Bernoulli process called $X[t]$. In this process, $P(X[t] = 1) = p$ and $P(X[t] = 0) = 1-p.$

We have a new process formed from $X[t]$ that is:

$$ Y[t] = \frac 12 \left(X[t] + X[t-1]\right) $$

I have to find the autocorrelation $r_Y [t+\tau,t]$. Here's what I got so far:

\begin{align} r_Y [t+\tau,t]& = E\left(Y[t+\tau]Y[t]\right)\\ &= \frac 14 E\left\{\left(X[t+\tau] + X[t+\tau-1]\right)\left(X[t] + X[t-1]\right)\right\}\\ &= \frac 14 \left(r_X[t+\tau,t]+r_X[t+\tau,t-1]+r_X[t+\tau-1,t]+r_X[t+\tau-1,t-1]\right) \end{align} From there, I can get $r_x[t+\tau,t]$:

$$ r_X[t+\tau,t] = \begin{cases} p^2 & \quad \text{if } \tau \neq 0\\ p & \quad \text{if } \tau = 0\\ \end{cases} $$

But the wall that I am running into (and maybe it's because it's not making sense to me because of lack of sleep) is how you substitute that expression back into $r_Y[t+\tau,t]$? I know you end up with three different answers, but trying to substitute it back in seems to cause a mess for me. I am hoping someone will help me with that part.

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  • $\begingroup$ For each substitution try grouping offsets with $\tau$ by substituting $t'=t-1$ (as one example). $\endgroup$
    – Peter K. ♦
    Commented Mar 31, 2016 at 0:39
  • $\begingroup$ what if you try to calculate the autocorrelation of $X(t)$ as a function of $Y(t)$ and then compare? if you are familiarize with the FT, you want to write $Y(t)$ as $Y(f)$ first, then find $X(f)$ as a function of $Y(f)$ and then calculate the autocorrelation of $X(f)$ and go again to the time domain. $\endgroup$
    – Behind The Sciences
    Commented Mar 31, 2016 at 6:22
  • $\begingroup$ Cross-posted to Engineering.SE. $\endgroup$
    – Chris Mueller
    Commented Mar 31, 2016 at 13:31

1 Answer 1

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It is helpful to realize that the auto-correlation of $X[t]$ does not depend on $t$, just on the time difference $\tau$. So we can write it as $r_X(\tau)$. Your result, which is correct, can then be written as

$$E(Y[t+\tau]Y[t])=\frac14\left(r_X(\tau)+r_X(\tau+1)+r_X(\tau-1)+r_X(\tau)\right)\tag{1}$$

Using the expression for $r_X(\tau)$, which is correctly given in your question, $(1)$ can be simplified to

$$E(Y[t+\tau]Y[t])=\begin{cases}\frac{p}{2}(1+p),&\tau=0\\ \frac{p}{4}(1+3p),&\tau\in\{-1,1\}\\p^2,& \text{otherwise}\end{cases}$$

which is of course also independent of $t$.

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  • $\begingroup$ I do not believe that his $r_x[\tau]$ is correct. Since $P(X[t])=0$ unless $t=0,1$, the autocorrelation should be $r_x[\tau]=0$ unless $\tau=0,\pm1$. $\endgroup$
    – Chris Mueller
    Commented Mar 31, 2016 at 13:45
  • $\begingroup$ @ChrisMueller Perhaps you are using autocovariance instead of autocorrelation? The OP's calculation of $r_x[t+\tau,t]$ is correct if we use the standard definition of autocorrelation as $E[X(t+\tau)X(t)]$: it is the autocovariance that has value $p(1-p)$ for $\tau=0$ and value $0$ for $\tau\neq 0$. $\endgroup$
    – Dilip Sarwate
    Commented Mar 31, 2016 at 14:46
  • $\begingroup$ @Dilip Ah, I see. I was using the continuous time definition: $\int_{-\infty}^\infty f(t)f(t+\tau)dt$. $\endgroup$
    – Chris Mueller
    Commented Mar 31, 2016 at 14:55

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