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I read this chapter about weakly stationary process from the book "Introduction to Probability, Statistics, and Random Processes" by Hossein Pishro-Nik.

Here is the defition of it: $$E[X(t_1)]=E[X(t_2)]$$ $$E[X(t_1)X(t_2)]=E[X(t_1+\Delta)X(t_2+\Delta)] = R_X(t_1,t_2)$$ The first equation is about equality of means and the second one is about equality of correlation function.

Then author writes that

The second condition states that the correlation function $R_X(t_1,t_2)$ is only a function of $\tau=t_1−t_2$, and not $t_1$ and $t_2$ individually.

I have a question about it.

How from $E[X(t_1+\Delta)X(t_2+\Delta)]$ follows that correlation depends only from $t_2 - t_1$?

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The equation holds for any $\Delta$. So you can set $\Delta= -t_1$. You would get $E[X(t_1) X(t_2)] = E[X(t_1 + \Delta) X(t_2 + \Delta)] = E[X(0) X(t_2-t_1)]$. The last expression depends only on $t_2-t_1$.

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    $\begingroup$ Good proof. but how to prove it in more general case? $\endgroup$
    – Kenenbek Arzymatov
    Commented Sep 3, 2021 at 14:43
  • $\begingroup$ This holds for and $t_1, t_2$. How do you want to generalize this more? $\endgroup$
    – Kota Mori
    Commented Sep 3, 2021 at 15:23
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    $\begingroup$ you proved it only for single case when $\Delta = -t_1$. It's a special case, not general. $\endgroup$
    – Kenenbek Arzymatov
    Commented Sep 3, 2021 at 15:34
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    $\begingroup$ Hi Kenenbek: There's nothing to prove. The statement is the definition of weak stationarity, namely that the autocovariance of the process is only dependent on the distance (in time) between the associated observations. So, $E(X_{t1} + \triangle))(X_{t2} + \triangle) = E(X_{t1}X_{t2})$ is the mathematical version of this statement. Since it holds for any value of $\triangle$, it holds in general. $\endgroup$
    – mlofton
    Commented Sep 3, 2021 at 22:05
  • $\begingroup$ Your goal is to show that "$E(X(t_1) X(t_2))$" can be written as a function only of $t_2-t_1$. You can show this by setting $\Delta = -t_1$. The reason why you are allowed to pick this arbitrary value for $\Delta$ is because by definition $\Delta$ can be any real number. $\endgroup$
    – Kota Mori
    Commented Sep 3, 2021 at 23:40

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