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I have a question about the fluctuation of autocorrelation of a signal due to signal's noise.

I have a signal defined in $-1\leq t \leq 1$ as the following: $V(t)=kt+R(t)$, where $R(t)$ is the random noise, and $k$ is non-negative.

Then, when I calculate the autocorrelation of $V(t)$, $G(\tau)=\int_{-1}^{1-\tau} V(t)V(t+\tau)dt$ (using the commend CorrelationFunction of Mathematica), I could observe that the fluctuation of $G(\tau)$ reduces as $k$ increases from zero. Could you explain the origin of this observation? Is it because as $k$ becomes larger, there is more autocorrelation and $R(t)$'s contribution to $G(\tau)$ is diminished?

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    $\begingroup$ might need to define how $R(t)$ is computed in your simulation or measurement (however $G(\tau)$ is computed). unless this is with pencil and paper (theoretical), i doubt either $V(t)$ or $R(t)$ or $G(\tau)$ are continuous time. if they are discrete time, then it would be useful to know how many samples times that $t$ is sliced up into between -1 and +1. $\endgroup$
    – robert bristow-johnson
    Sep 24, 2018 at 4:31
  • $\begingroup$ $R(t)$ is discontinuous in time and $R(t)$ is sampled 8000 times(uniformly distributed in $t$). $R(t)$ is generated by RandomReal of Mathematica (RandomReal chooses reals with a uniform probability distribution). $\endgroup$
    – user16308
    Sep 24, 2018 at 14:02

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If you view $V(t)$ as desired signal $kt$ affected by noise $R(t)$, the clearly as you keep increasing $k$, this has the effect of increasing signal energy. Hence if the noise power is kept constant then $V(t)$ tends towards being more deterministic (in a loose sense) with improving SNR, and autocorrelation hence is more deterministic with reduced fluctuations.

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It is necessary to de-trend your signal first subtracting off the $kt$. Otherwise, you will be subjected to what is called the "spurious regression effect" in econometrics and your autocorrelation results won't make any sense. See the link below for an explanation of spurious regression. IIRC, the explanation discusses the effect in terms of "difference stationary" processes but the same is true for what you have which is called a "trend stationary" process. So, just de-trend your series and then do what you just did and the result should be fine and the autocorrelation should decrease with greater lag.

P.S: This assumes that $R(t)$ is not white noise. Otherwise, if it is, then there is obviously no autocorrelation in your process once you detrend.

https://wolfweb.unr.edu/homepage/zal/STAT758/Granger_Newbold_1974.pdf

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  • $\begingroup$ white noise cannot really exist in a physical manner because it has infinite power. in uniformly sampled-systems, bandlimited white noise (white up to Nyquist) is what comes outa a good random number generator. most random number generators will spit out uniform distribution virtually white noise and you can either map or add up a few uniform random variables to get a virtually gaussian white noise. but otherwise, i dunno how the OP is looking at $R(t)$. $\endgroup$
    – robert bristow-johnson
    Sep 24, 2018 at 4:26
  • $\begingroup$ thanks robert. good point. I should have said "independent" rather than white noise. Either way, if that's the case, then the empirical autocorrelation should be close to zero. There are statistical procedures for testing if it's zero. ( box-leung, portmanteau etc ). $\endgroup$
    – mark leeds
    Sep 24, 2018 at 5:52
  • $\begingroup$ Someone experimentally measured $V(t)$ and obtained $G(\tau)$. And I want to investigate whether there is a trend in $V(t)$, by looking at the obtained $G(\tau)$. I can see some random fluctuation in $G(\tau)$, which I can see only from a simulation of small $|k|$. So it seems like that the $k$ for the experiment is approximately zero, but I want to double check this. $\endgroup$
    – user16308
    Sep 24, 2018 at 13:48
  • $\begingroup$ Hi: I'm not clear on whether $R(t)$ is independent or not ? If it is, then it's best to google for "testing for trend" or "trend stationary" ( because there is no autocorrelation ) . If it isn't, then you've got a more complex problem that I'd have to sit down and think about more and I don't have time at the moment. Hopefully someone else can comment or answer in the latter case. $\endgroup$
    – mark leeds
    Sep 24, 2018 at 16:10
  • $\begingroup$ $R(t)$ is independent. Thank you for your answers. $\endgroup$
    – user16308
    Sep 24, 2018 at 17:12

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