1
$\begingroup$

I have 16 NDVI values for 16 years. I want to use this time series data set to investigate a trend in the dataset to distinguish between noise, randomness and actual trends.

Can I use R to do this? Or is there any other trend software for this? I just want a method that is a bit simple to understand and run as a beginner.

$\endgroup$
  • $\begingroup$ There are multiple techniques, but if you just want to get your hands dirty try running an AR(1) or MA(1) process. There are a bunch of concepts like stationarity and ergodicity and order p,q but if you just want to do something simple run those $\endgroup$ – user2879934 May 1 '17 at 17:35
0
$\begingroup$

In order to distinguish between between random noise and actual trends it is necessary to have a model for the noise itself. For example, if the noise is Poisson type, which occurs in various counting situations like radioactive decay counting, then for $n$ counts, one expects $\sqrt{n}$ as a single standard deviation of counting error. If one expects Gaussian noise, one must have a numerical expectation of what that noise is. Subsequently, one can calculate what the fit error from the model alone is. Here is an example from EJNMMI-physics:

"Assuming counting error of a Poisson noise type, misregistration, an imaging term is defined graphically as the standard deviation vertical misalignment on a square root of count rate TAC plot, which latter is, indeed, an image. Misregistration standard deviation, $σ_M = 0.79943\%$, was calculated from the well-known equation for correlated variances $$σ_M^2=σ_{F,N}^2=σ_F^2−σ_N^2−2ρ_{F,N}σ_Fσ_N,\text{ }\text{ }\text{ }\text{ }\text{ } (11)$$ where the variance of misregistration, $σ_M^2$, is $σ_{F,N}^2$, the variance of the difference between fit error, $σ_F = 1.44164\%$ [from the standard deviation from Table 1 of Eq. (10)], and noise error, $σ_N = 1.02864\%$ [Eq. (8)], where $ρ_{F,N} = 0.76531$ is the correlation coefficient of $F$ and $N$."

What programming language you use is your preference. Moreover, without more physical information about the noise characteristics of your measurement system, the question cannot be answered at all. The question asked is rather insightful. The answer, if any, requires more information and may be challenging for a beginner.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.