My understanding is Maximum Likelihood and DFT Peak Finding for a single tone produce the same results assuming the ML is restricted to the same frequencies as the DFT.

I was wondering if there was an easy way to prove this?
My understanding is that ML will be the same doing Least Squares regression so I suppose an equivalent statement would be to prove that the DFT peak gives the best Least Squares estimate of single tone sinusoidal.

  • $\begingroup$ I think I figured it out: DFT applies a correlation, choosing the strongest correlation(peak detection) gives the best estimate using the difference squared norm which is equivalent to minimizing the sum in joint PDF for uncorrelated gaussians. Am I wrong? $\endgroup$ – FourierFlux Feb 9 '19 at 20:30
  • $\begingroup$ yea sometimes I forget. $\endgroup$ – FourierFlux Jun 27 '19 at 7:02

Maximum Likelihood under the assumption of Additive White Gaussian Noise (AWGN) is always equivalent to finding the hypothesis with the minimum distance to given data.

Since minimizing distance is equivalent (In the euclidean Space) of maximizing the correlation you can always build the idea of Match Filter for parameter estimation in the settings of ML under AWGN.

Matched filter means that you create a set of hypothesis parameterized on a grid and goes one by one until you find the one which maximizes correlation (Minimizes distance) to data.

This is exactly what's the DFT is doing.
The grid is based on the different tones (Frequencies) and the correlation level is given by the bin value.

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