Proof of Theorem 8.
We consider the case of continuous σ; the proof of the discrete case is similar. Taking negative logs on both sides of and using the power series representation − log(1 − x) = x + x2/2 + x3/3 + …, we obtain
Thus, we wish to show that, for suitable an and bn, for any x, while as n → ∞:
By the von Mises condition, as n → ∞, so we want
Now specializing to Fi(x) = exp[−(σi/xρ)] = exp(−τi/xρ), with we have φi(x) =τix− ρ, φ′i (x) = − ρτi x− (ρ + 1), and ψi(x) = xρ + 1/ρτi, so that
Letting u = i/n (i = 1, …, n) and assuming there exists a continuous ...