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I have a question about how a result for the auto-correlation was calculated.

The autocorrelation function $R_X(t_1,t_2)$ is defined as $R_X(t_1,t_2) = E\left[X(t_1)X(t_2)\right]$

The stochastic process $X(t)$ is characterized by: $ X(t) = \begin{cases} +\sin t, \ \ p=1/4\\ -\sin t, \ \ p=1/4\\ +\cos t, \ \ p=1/4\\ -\cos t, \ \ p=1/4 \end{cases}$

The autocorrelation function result is: $R_X(t_1, t_2) = \frac12 \cos(t_2 - t_1)$. How was this reached? Specifically, I had trouble with applying the expectation over two different time instances.

Thanks, I appreciate any help.

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    Hi Didier, I had a feeling the process $X(t)$ was independent for different values of $t$. But when I calculated the expectation, the 16 terms all cancelled each other out and the autocorrelation was 0.2012-03-12

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As indicated in the comments, the autocorrelation function describes (an aspect of) the stochastic structure of a process, namely, its two-dimensional marginals. Hence, the stochastic process $(X(t))_t$ is not characterized by the distribution of each $X(t)$ and, unless one adds a description of the dependence between $X(t_1)$ and $X(t_2)$ for every $t_1\ne t_2$, the question is not answerable.

Here is a possible dependence structure. Let $U$ denote a random variable uniformly distributed on $\{0,1,2,3\}$. Define $X(t)=\sin(t)$ if $U=0$, $X(t)=\cos(t)$ if $U=1$, $X(t)=-\sin(t)$ if $U=2$ and $X(t)=-\cos(t)$ if $U=3$. Then the marginal of $X(t)$ is as desired and one may group these definitions into the condition that $ X(t)=\sin\left(t+\frac\pi2 U\right). $ Note that the same $U$ is used to define $X(t)$ for every $t$.

For every $t$ and $s$, $\mathrm E(X(t))=\mathrm E(X(s))=0$ and $\mathrm E(X(t)X(s))$ is $ \frac14\left(\sin(t)\sin(s)+\cos(t)\cos(s)+(-\sin(t))(-\sin(s))+(-\cos(t))(-\cos(s))\right), $ which is $\tfrac12\cos(t-s)$, hence $ R_X(t,s)=\mathrm E(X(t)X(s))-\mathrm E(X(t))\mathrm E(X(s))=\frac12\cos(t-s). $