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Question

Solution of the question

The Solution is given above:

The Question is, how did the $\mathbb{E}{[x(k)f(l)]}$ and $\mathbb{E}{[x(l)f(k)]}$ become zero? is there some rule that correlation between Random Process and Deterministic sequence is zero?

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  • $\begingroup$ In a sea of poorly asked "help with my homework" questions, this one is done well. Congrats, and thank you. $\endgroup$
    – TimWescott
    Oct 17, 2022 at 19:41
  • $\begingroup$ Note that $E\{x(k)f(l)\}$ is the plain old correlation between them -- autocorrelation in the correlation of a signal with itself. $\endgroup$
    – TimWescott
    Oct 17, 2022 at 19:44

2 Answers 2

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Since $f(n)$ is deterministic, we have

$$E\{x(m)f(n)\}=f(n)E\{x(m)\}\tag{1}$$

So the value of $E\{x(m)f(n)\}$ is simply $f(n)$ times the mean of $x(n)$, and since the mean of $x(n)$ is zero, so is $E\{x(m)f(n)\}$.

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  • $\begingroup$ Even though Tim’s answer is also valid, this looks like it should be the accepted answer. $\endgroup$
    – Jdip
    Oct 17, 2022 at 21:00
  • $\begingroup$ yes, this makes sense mathematically. $\endgroup$ Oct 17, 2022 at 21:06
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$E\{x(k)f(l)\} = 0$ because $x(n)$ and $f(n)$ are uncorrelated.

Unless otherwise stated, in your coursework you can assume that any given random process is uncorrelated with another, or with a deterministic signal. The only reason that a signal and a random process may be correlated is because there's some causal link between them -- i.e., the deterministic signal is being generated in part by the random signal, or the random signal is being generated in part by the deterministic one.

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    $\begingroup$ The only reason why $E\{x(m)f(n)\}=0$ is because $E\{x(n)\}=0$. If $E\{x(n)\}\neq 0$ then also $E\{x(m)f(n)\}\neq 0$ (unless $f(n)=0$). $\endgroup$
    – Matt L.
    Oct 17, 2022 at 20:23
  • $\begingroup$ Please explain how that is true in the case where $x(m) = \sin m$ and $f(m) = \sin m$. $\endgroup$
    – TimWescott
    Oct 17, 2022 at 23:24
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    $\begingroup$ $x(n)$ is a random signal, so it can't be $\sin(n)$. It could be $\sin(n+\phi)$ with some random phase $\phi$. If that random phase has a uniform distribution then you would have $E\{x(n)\}=0$ and, consequently, $E\{x(m)f(n)\}=0$. $\endgroup$
    – Matt L.
    Oct 18, 2022 at 6:21

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