I’m looking to analytically calculate a probability distribution of sampling

points from an oscillating function when there is some measurement error. I have

already calculated the probability distribution for the “without noise” part (I will put this at the end), but I can’t figure out how to include “noise”.## Numerical estimate

To be clearer, imagine there is some function y(x) = \sin(x) which you randomly pick points from during a single cycle; if you bin the points in a histogram you will get something related to the distribution.

## Without noise

For example here is the sin(x) and the corresponding histogram

## With noise

Now if there is some measurement error then it will change the shape of the histogram (and hence I think the underlying distribution). For example

## Analytic Calculation

So hopefully I’ve convinced you there is some difference between the two, now I will write out how I calculated the “without noise” case:

## Without noise

y(x) = \sin(x)

Then if the times at which we sample are uniformly distributed then the probability distribution for y must satisfy:

P(y) dy = \frac{dx}{2\pi}

then since

\frac{dx}{dy} = \frac{d}{dy}\left(\arcsin(y)\right) = \frac{1}{\sqrt{1 – y^{2}}}

and so

P(y) = \frac{1}{2\pi\sqrt{1 – y^{2}}}

which with appropriate normalisation fits the histogram generated in the “no noise” case.

## With noise

So my question is: how can I analytically include noise in the distribution? I think it is something like combining the distributions in a clever way, or including noise in the definition of y(x), but I’m out of ideas and ways to move forwards so any hints/tips or even recommended reading will be much appreciated.

**Answer**

It depends on how the noise process is structured.

Assuming that I’ve understood your situation correctly, if the noise is additive, independent and identically distributed, you would just take the *convolution* of the noise density with the density of Y.

If X_i is random uniform over a cycle, your noiseless process conditional on x is Y_i|X_i=x_i, which is degenerate, with mean \sin(x_i) and variance 0. The marginal distribution of Y is a uniform mixture of those degenerate distributions; it looks like you have worked that distribution out correctly; let’s call that density g.

If, for example your noise is \epsilon_i\sim N(0,\sigma^2), which is to say f(\epsilon)=\frac{1}{\sqrt{2\pi}\,\sigma}\exp({\frac{\epsilon^2}{2\sigma^2}}), then f*g is the density of the sum of the noise with that uniform mixture of noiseless variables.

f_{Y+\epsilon}(z) = (f*g)(z) = \int_{-\infty}^{\infty} f_{Y}(y)f_{\epsilon}(z-y)dy=\int_{-\infty}^{\infty} f_{Y}(z-w)f_{\epsilon}(w)dw

(this convolution was done numerically; I don’t know how tractable that integral is in this example, because I didn’t attempt it.)

**Attribution***Source : Link , Question Author : Greg , Answer Author : Glen_b*