Pareto exceedance probability by invariance

Question

Excess claims above a known floor xm are Pareto with unknown index \alpha, density f(x \mid \alpha) = \dfrac{\alpha xm^{\alpha}}{x^{\alpha+1}} for x \ge xm. A sample of n = 90 claims has \sumi \ln(xi / xm) = 30. What is the maximum likelihood estimate of the probability that a claim exceeds 2 xm?

Accepted Answer

First the tail index. The MLE is \hat\alpha = \frac{n}{\sumi \ln(xi / xm)} = \frac{90}{30} = 3. For a Pareto, the exceedance probability is P(X t) = (xm / t)^{\alpha}. At t = 2xm this is (1/2)^{\alpha}, a function of \alpha. By invariance, plug in \hat\alpha: \widehat{P(X 2xm)} = \left(	frac{1}{2}ight)^{\hat\alpha} = \left(	frac{1}{2}ight)^{3} = \frac{1}{8} = 0.125. Takeaway. Estimate the index, then read any tail probability off the Pareto survival function using MLE invariance.

Pareto exceedance probability by invariance

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