Or is it maybe the same thing? I see that autocorrelation is when Yt is correlated with its lag Yt-1. But isn't that essentially what an AR process (say AR(1)) is? We are assuming that there IS correlation between its previous time period since we might see a directional trend from its initial data, right?

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    $\begingroup$ An autocorrelation is a function of a pair of times. An AR process is a stochastic process: thus, they aren't even remotely the same kinds of mathematical objects. On the face of it, then, your question makes no mathematical sense. Could you clarify what you're trying to ask? $\endgroup$ – whuber May 12 '19 at 19:53

Think of it this way:

If the index $t$ denotes time, then a stochastic process is simply a collection of random variables indexed by time.

There are stochastic processes (e.g., AR(1)) for which we can explicitly indicate how the value of $Y$ at time t (that is, $Y_t$) depends on values of Y at previous times. For example, an AR(1) process with mean zero stipulates that $Y_t = \phi*Y_{t-1}+ \epsilon_t$, where $\epsilon_t$ is a white noise process with zero mean and constant variance $\sigma^2$.

The autocorrelation function provides a measure of similarity between the values of $Y$ at times $t$ and $s$ by computing the correlation between $Y_t$ and $Y_s$, namely $Cor(Y_t, Y_s)$. (For a weekly stationary stochastic process, this correlation only depends on how apart in time $t$ and $s$ are from each other, it does not depend on the actual values of $t$ and $s$.)

For the above AR(1) process, the correlation between $Y_t$ and $Y_{t-1}$ is given by:

$Cor(Y_t, Y_{t-1}) = Cor( \phi*Y_{t-1}+ \epsilon_t, Y_{t-1}) = \phi$

$Cor(Y_t, Y_{t-2}) = \phi^2$

and, more generally,

$Cor(Y_t, Y_{t-k}) = \phi^k$, where $k > 2$ is an integer. (See https://www.mathstat.dal.ca/~stat5390/Section_3_ACF.pdf.)

  • $\begingroup$ Your opening characterization of stochastic processes seems quite at odds with standard conceptions of them. Did you perhaps mean to describe AR models? $\endgroup$ – whuber May 12 '19 at 21:17
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    $\begingroup$ Thanks, @whuber! I made some edits to the original answer to reflect your concerns - let me know if they look fine to you. I want to keep things simple. $\endgroup$ – Isabella Ghement May 12 '19 at 21:24

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