1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ For each substitution try grouping offsets with $\tau$ by substituting $t'=t-1$ (as one example). $\endgroup$ – Peter K. Mar 31 '16 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 Mar 31 '16 at 6:22
  • $\begingroup$ Cross-posted to Engineering.SE. $\endgroup$ – Chris Mueller Mar 31 '16 at 13:31
3
$\begingroup$

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 $t$.

$\endgroup$
  • $\begingroup$ 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 Mar 31 '16 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 Mar 31 '16 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 Mar 31 '16 at 14:55

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.