On the distribution of burr with applications

Abstract

Under certain conditions, the distribution of burr is shown to follow an extreme value distribution. In this context, a result on extremal process based on stationary sequence is proved. Some data sets are analyzed and applications of the results indicated.

Appendix

Here we prove Theorem 2, extending A3 of Dasgupta et al. (1981). The earlier proof of Dasgupta et al. (1981) has to be suitably modified so as to extend the stated results from ‘iid continuous random variables’ to ‘stationary random variables’. We provide a modified proof elaborated below.

Assume that the distribution of the stationary sequence X _i ’s are nondegenerate as the proof is trivial otherwise.

Consider a fixed sequence i _o  = i _o (m) = o(m). If the value of max _{1 ≤ j ≤ m} X _j is attained at a single index of X, then from stationarity of the random variables, $P \{\cup_{i = 1}^{i_o}( X_i = \max_{1\leq j\leq m} X_j)\} = \frac{i_o}{m};$ since the index of X, attaining the maximum value of X, is uniformly distributed on the set {1, ⋯ ,m}.

In a similar fashion, if the maximum is attained for two distinct values of the index, then the probability that both the indices are in the set {1, ⋯ ,i _o }, is $\frac{i_o(i_o-1)}{m (m-1)}\leq \frac{i_o}{m},$ etc.

Thus, in general, the probability that at least one index with maximum value of X will lie within the set { i _o  + 1, ⋯ ,m } is at least $(1 - \frac{i_o}{m}).$ In other words,

$$ P \{\cup_{i = i_o +1}^m( X_i = \max\nolimits_{1\leq j\leq m} X_j)\}\geq 1 - \frac{i_o}{m} \rightarrow 1, \;\mbox{as}\; m \rightarrow \infty. $$

Now using the fact that {c _i } is a nondecreasing sequence, one gets

$$ P \{\cup_{i = i_o +1}^m( c_i X_i = \max\nolimits_{1\leq j\leq m} c_j X_j)\}\geq 1 - \frac{i_o}{m} \rightarrow 1. $$

On the intersection of above two sets, we have from Eq. 4.1

$$ c_{i_o}\max\nolimits_{1\leq i\leq m} X_i\leq Y_m \leq \max\nolimits_{1\leq i\leq m} X_i. $$

Thus it suffices to show that

• (1) $b_m^{-1}(1 - c_{i_o}) \max_{1\leq i\leq m} X_i\rightarrow 0,\;\;\; \mbox {in distribution}.$

  Now from Eq. 4.2, $Z_m = b_m^{-1}( \max_{1\leq i\leq m} X_i - a_m)$ is bounded in probability. Write,

  $$ b_m^{-1}(1 - c_{i_o}) \max\nolimits_{1\leq i\leq m} X_i = (1 - c_{i_o})(a_m b_m ^{-1}+ Z_m). $$

  Since $c_{i_o}\rightarrow 1,$ it is sufficient to show that,

  $$ (1 - c_{i_o})|a_m b_m ^{-1}|\rightarrow 0, $$

  i.e., $c(i_o(m)) = 1- o(|\;a_m^{-1}b_m|).\;\;$ Hence the first part of the theorem.

  To prove the second part, note that (1 − c(k))f(k) = o (1), there exists a sequence h(k)→ ∞ such that

• (2) (1 − c(k))f(k)h(k) = o (1). (e.g., if ((1 − c(k))f(k) ≤ ε _k →0, then one may take $h(k)= \epsilon_k^{-1/2}.)$

  Let n = n(k) be such that

• (3) f(n(k)) < f(k − 1)h(k − 1) ≤ f(n(k) + 1).

  Such a choice of n = n(k) is possible as $f(i) \rightarrow \infty, \;\mbox {as}\;i \rightarrow \infty.$

  (For a fixed k the middle term of the above is finite and as f(n(k))→ ∞ , n(k)→ ∞ ; (3) is ensured.)

  From the assumption $\overline{\lim}_{i \rightarrow \infty}\frac{f(i\delta)}{f(i)} < \infty,$ for every fixed δ > 0, it follows that

• (4) k^− 1n(k)→ ∞ .

  To see this, suppose n(k) < < k, divide the terms of (3) by f(k); then both r.h.s. and l.h.s. terms of (3) are bounded above in view of $\overline{\lim}_{i \rightarrow \infty}\frac{f(i\delta)}{f(i)} < \infty,$ whereas the middle term → ∞ , leading to a contradiction.

  Next, for m = 1, 2, ⋯ define, k(m) = k if n(k) ≤ m < n(k + 1). Then

• (5) f(m) < f(n(k + 1)) < f(k)h(k) see (3). Now from (2)

• (6) (1 − c _k(m))f(m)→0, where $\frac{k(m)}{m}\leq \frac{k(m)}{n(k(m))}\rightarrow 0,$ from (4).

Hence the second part.

We have a different bound for burr Y given in Eq. 4.6. Proof of Theorem 2 remains valid for every fixed j = 1, ⋯ ,p ; as the sequence of constants {c _1j , ⋯ ,c _mj } satisfy the conditions therein. Theorem 2 therefore, holds for the burr Y satisfying the bound of Eq. 4.6, as the minimum is taken over j, a finite p number of terms.