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Suppose we have a time series $x_t=\sin(0.02\pi t)$. Although this time series is totally deterministic, we can treat it as one realization of a proto/quasi/pseudo-stochastic process and estimate the distribution of $P(x_t)$. For example, if kernel density kernel is used, we will get something like the following image enter image description here Obviously the estimations at the two ends of the interval [-1,1] are messed up since a sine series can never go outside this range. The image just gives you an idea of what I want.

Question: What is the analytical expression of $P(x_t)$?

Edit 1: The bandwidth I used is the Silverman's rule: B.W. Silverman, “Density Estimation for Statistics and Data Analysis”, Vol. 26, Monographs on Statistics and Applied Probability, Chapman and Hall, London, 1986.

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  • $\begingroup$ You might like to vary the kernel width for the KDE plot (I assume this is with the Gaussian kernel, but that is not important), narrower would be interesting $\endgroup$ – Conrad Turner Dec 21 '14 at 17:05
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Suppose we sample uniformly from 1 cycle of the sine wave, then it is quite clear (I would suggest drawing a picture) that the cumulative distribution function is: $$ F(x)=\frac{\arcsin(x)}{\pi}+\frac{1}{2},\ \ \textrm{where}\ -1\le x\le1 $$ A longer explanation of this can be found here.

Then density can then be found by differentiating this to get: $$ f(x)=\frac{1}{\pi\sqrt{1-x^2}},\ \ \textrm{where}\ -1\le x\le1 $$ and looks like:

enter image description here

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  • $\begingroup$ I have no idea why your answer got downvoted... by the way, since you commented on the width of the kernel, I would like to know if you have tricks to make KDE deal with this kind of probability distribution (prob. dist. with finite support) correctly, other than decreasing the width of the Gaussian kernel. $\endgroup$ – wdg Dec 25 '14 at 8:03
  • $\begingroup$ I don't have any trick foe use with KDE, it is only an approximation scheme and there are no kernels that will reproduce the singularities at the end of the interval that we see here. The real solution is to recognise that something odd will happen when you have a distribution with interval support and transform the the interval into the entire real line. $\endgroup$ – Conrad Turner Dec 25 '14 at 9:47

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