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Assuming that $S_X(w)$ denotes powers spectral density function at frequency $w$, we are given $$S_X\left(\frac{\pi}{4}\right)=10+3\sqrt{2},\quad S_X\left(\frac{\pi}{6}\right)=11+3\sqrt{3}.$$ We also know that the autocorrelation function $R_X$ has the properties $R_X(0) = 10$ and $R_X(m) = 0$ for $|m|\geq 3$. What is the value of $\frac{R_X(1)}{R_X(2)} $?

An attempt: It is known that $\lim_{m\rightarrow \infty}R_X(m) = E^2\{X\}$. Since $R_X$ is assumed to be $0$ for large $m$, we immediately conclude that $E^2\{X\}=0$.

On the other hand the autocorrelation can be obtained from $S_X$ by applying the inverse Fourier transform. This would lead to $$ R_X(1) = \int_{-\infty}^{+\infty} S_X(f) e^{2i\pi f }\,\mathrm d\!f$$ $$ R_X(2) = \int_{-\infty}^{+\infty} S_X(f) e^{4i\pi f }\,\mathrm d\!f$$ but calculating these integrals look formidable as is. Maybe we can use our information on $E^2\{X\}$? Anyhow, I have no clue on how to continue. Your help is greatly appreciated!

(If it's any help, the solution manual says $\frac{R_X(1)}{R_X(2)}=3$, but it doesn't explain how or why.)

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    $\begingroup$ According to the properties of autocorrelation function, $R_X(0)\geq |R_X(k)|$ for all $k$, so how is it $0$? $\endgroup$ – gunes Feb 14 at 0:14
  • $\begingroup$ My bad. It's $R_X(0) = 10$. I made a typo. Do you think I have to repost the question with the correction? I guess a lot of the viewers may have moved on because the problem couldn't be solved because of the typo. $\endgroup$ – User32563 Feb 14 at 10:14
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I think there are several things to assume further in the question. The given equations are probably for $\omega=\pi/4$ and $\omega=\pi/6$, instead of $f=\pi/4$ and $f=\pi/6$ for the ease of calculations. $S_x(\omega)$ is another and common way of writing PSDs, especially when the given input series is discrete (although it's not stated explicitly in the question). Therefore, I'm going to take the DTFT of the discrete autocorrelation series to obtain the PSD:

$$S_x(\omega)=\sum_{k=-3}^3 R_x(k)e^{-jwk}=R_x(0)+2R_x(1)\cos (\omega)+2R_x(2)\cos (2\omega)+2R_x(3)\cos(3\omega)$$

And, substituting frequencies in the given equations, we have: $$S_x(\pi/4)=10+R_x(1)\sqrt 2-R_x(3)\sqrt 2=10+3\sqrt 2$$ $$S_x(\pi/6)=10+R_x(1)\sqrt 3+R_x(2)=11+3\sqrt 3$$

It's not possible to solve this uniquely. So, another assumption of mine is $R_x(3)=0$, i.e. some another typo in the question statement that $R_x(k)=0$ for $|k|\geq 3$ instead of $|k|>3$, using these equations, we'll find that $R_x(1)=3,R_x(2)=1$, so the answer would be correct.

Notes on the literature difference: Many sources in the time-series literature defines the autocorrelation function a bit differently than the signal processing literature. Here, I believe it's assumed that $R_x(k)=E[x_nx_{n-k}]$, instead of $\rho(k)=\operatorname{cov}(x_n,x_{n-k})/\operatorname{var}(x_n)$.

Notes to the OP: I've made a lot of assumptions to correct the question statement, which couldn't be possible if I weren't quite familiar with the topic. So, either the question statement has full of errors in it, or you mistyped a lot of things. This highly decreases the chance of someone answering your question, so I advise you to do your best while asking questions.

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    $\begingroup$ Many thanks for the insightful solution and corrections. You were absolutely right about the typos, I should have double-checked the details. Also this problem Indeed comes from signal processing, your assumption on $R$ is also correct. $\endgroup$ – User32563 Feb 14 at 16:32
  • $\begingroup$ I just got a small question, if you humor me. Have you also assumed that $R_x(m) = R_x(-m)$? According to my textbook, this is valid for stationary processes, is it also valid here in your opinion? $\endgroup$ – User32563 Feb 14 at 19:16
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    $\begingroup$ Yes, I assumed that as well. $\endgroup$ – gunes Feb 14 at 19:18

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