MLE for the scale of Rayleigh fading

Question

The envelope amplitude of a fading radio signal (in millivolts) is modeled as Rayleigh with unknown scale $\sigma$ , density

$f(x \mid \sigma) = \frac{x}{\sigma^2}\, e^{-x^2/(2\sigma^2)}, \qquad x > 0.$

Four amplitudes are measured: 2, 4, 4, 6.

Derive the maximum likelihood estimator of $\sigma$ and compute it.

Accepted Answer

The log of one density is \ln f = \ln x - 2\ln\sigma - \dfrac{x^2}{2\sigma^2}. Summing and keeping the \sigma-dependent terms, \ell(\sigma) = -2n\ln\sigma - \frac{1}{2\sigma^2}\sumi xi^2 + 	ext{const}. It is easier to differentiate with respect to \sigma^2. Writing v = \sigma^2, \ell(v) = -n\ln v - \frac{1}{2v}\sumi xi^2, \qquad \ell'(v) = -\frac{n}{v} + \frac{1}{2v^2}\sumi xi^2 = 0. Solving gives \hat\sigma^2 = \frac{1}{2n}\sumi xi^2, \qquad \hat\sigma = \sqrt{\frac{\sumi xi^2}{2n}}. Here \sumi xi^2 = 4 + 16 + 16 + 36 = 72 and n = 4, so \hat\sigma^2 = \frac{72}{8} = 9, \qquad \hat\sigma = 3…

MLE for the scale of Rayleigh fading

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