Monday, 17 August 2009

light - how to assess a periodical signal?

You have it the wrong way around. You have to establish a periodicity before doing the folding.



There are various ways to do this, but the most frequently (pun intended) used is the Lomb-Scargle periodogram.



Start with a datset consisting of a set of points $(t_i, m_ipm sigma_i )$, where $t_i$ are times of observation and $m_i$ and $sigma_i$ are the fluxes and their uncertainties at those times.



You then calculate an array of chi-squared statistics, by fitting a mean level $bar{m}$ and sinusoidal functions over a range of discrete frequencies $omega$. The aim is to maximise the statistic
$$p(omega) = frac{chi^2_{0} - chi^2(omega)}{chi^2_{0}},$$
where
$$chi^{2}_{0} = sum (m_i - bar{m})^2/sigma_i^{2}$$ and
$$chi^{2}(omega) = sum (y_i - bar{m})^2/sigma_i^{2},$$ with $y_i = Asin(omega t_i + phi) + A_0$, where $A$, $A_0$ and $phi$ are free parameters that are optimised for each frequency (i.e. you find the minimum possible $chi^2(omega)$ at each frequency).



Once a period is obtained from a peak in the power spectrum, $p_{best}$, then it must be established whether this peak is "real" or significant.



Theoretically, the probability that $p>p_{best}$ is $(1-p_{best})^{(N-3)/2}$ for $N$ independent measurements (see Cumming, Marcy & Butler 1999). This is the "false alarm probability" - the chance that you would get $p>p_{best}$ from random, aperiodic data with those uncertainties.



However, real data has correlated noise and the points are not independent. So what you do instead (see Collier Cameron et al. 2009) is shuffle all the times of the data points (i.e. randomly reallocate the times of each observation) and go through the process again. Do this a squillion times with different random reallocations and see how many times you end up with $p>p_{best}$. This gives you some idea about how confident you can be that your detected periodic signal is real. This is called a "bootstrap Monte-Carlo method".



There are other ways to do this; my answer is not complete; but it's the one I'm familiar with.

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