MLE for the tail index of large losses

Question

Operational losses above a known floor xm = 1	ext{M} are modeled as Pareto with unknown tail index \alpha, density f(x \mid \alpha) = \frac{\alpha\, xm^{\alpha}}{x^{\alpha+1}}, \qquad x \ge xm. A sample of n = 100 excess losses has \sumi \ln(xi / xm) = 40. Derive the maximum likelihood estimator of \alpha and compute it.

Accepted Answer

The log of one density is \ln f = \ln\alpha + \alpha\ln xm - (\alpha+1)\ln x. Summing over the sample, \ell(\alpha) = n\ln\alpha + n\alpha\ln xm - (\alpha+1)\sumi \ln xi. Differentiate with respect to \alpha: \ell'(\alpha) = \frac{n}{\alpha} + n\ln xm - \sumi \ln xi = \frac{n}{\alpha} - \sumi \ln\frac{xi}{xm} = 0. Solving, \hat\alpha = \frac{n}{\sumi \ln(xi / xm)}. The data enter only through the sum of log-ratios. Here that sum is 40 with n = 100, so \hat\alpha = \frac{100}{40} = 2.5. Takeaway. The Pareto tail index MLE is the count divided by the sum of log-excesses \ln(xi/xm); a bigger \ha…

MLE for the tail index of large losses

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