Losses above a threshold u = 1 follow a Pareto tail with P(X > x) = (x/u)^{-alpha} for x > u. From 50 exceedances whose values of log(x_i/u) sum to 25, derive the maximum likelihood estimator of alpha, compute it, and give its approximate standard error.

Losses above a threshold u = 1 follow a Pareto tail with P(X > x) = (x/u)^{-alpha} for x > u. From 50 exceedances whose values of log(x_i/u) sum to 25, derive the maximum likelihood estimator of alpha, compute it, and give its approximate standard error.

Approach: Reparameterise in terms of the log-exceedances, which are exponential, then maximise the resulting one-parameter log-likelihood and read the information off it.

2. The density is f(x) = alpha*u^alpha*x^{-(alpha+1)}, so in terms of y_i = log(x_i/u) the log-likelihood is n*log(alpha) - alpha*sum(y_i). Setting the derivative n/alpha - sum(y_i) to zero gives alpha_hat = n/sum(y_i) = 50/25 = 2, which is the Hill estimator of the tail index. The y_i are iid Exponential(alpha), so the Fisher information per observation is 1/alpha^2 and the asymptotic standard error is alpha/sqrt(n) = 2/sqrt(50) = 0.283. A 95% interval runs from 1.45 to 2.55. The second moment of the loss exists only for alpha > 2, and that interval straddles 2, so 50 exceedances cannot settle whether the variance of the loss is finite, which is the parameter a risk limit actually depends on.

Follow-up: Lowering the threshold u brings more exceedances into the estimate. What is the bias and variance trade-off in that choice?

Key concepts: Hill estimator, maximum likelihood, Fisher information, tail index.