By Moti L. Tiku
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Example text
3) virtually equal to zero even for small sample sizes. This establishes the fact that the ML and the MML estimates are numerically the same (almost). 137), Tan (1985), Tiku and Vaughan (1997, pp. 101, 106-107) and Vaughan (2002) arrive at the same conclusion. Also, the MML estimators are asymptotically fully efficient and are highly efficient for small sample sizes. Smith et al. (1973), Lee et al. (1980), Tan (1985), Vaughan (1992a, 1994) and Senoglu and Tiku (2001) have similar results. There is, therefore, no reason whatsoever for not using the MML estimators in place of the elusive ML estimators.
PM5 18 Robust Estimation and Hypothesis Testing APPENDIX 1A EXPECTED VALUES AND VARIANCES AND COVARIANCES Let y(1) ≤ y(2) ≤ ... ≤ y(n) be the order statistics of a random sample of size n from a distribution of the type (1/σ)f ((y – µ)/σ). Write z = (y – µ)/σ and z(r) = {y(r) – µ}/σ, 1 ≤ r ≤ n. 9) n! [F(u)]r–1 [1 – F(u)]n–r f(u), – ∞ < u < ∞, (r − 1) ! (n − r ) ! 2) f (z)dz is the cumulative distribution function. The joint probability density function of u = z(r) and v = z(s) (r < s) is given by (David, 1981, p.
1). Now, d ln L n 1 =− + 3 dσ σ σ n ∑ (y i i=1 − µ ) 2 = 0. n S2 = Writing ∑ (y i − µ) 2 / n i=1 d ln L n = 3 (S – σ) (S + σ) dσ σ n = 3 (S2 – σ2) = 0. 8) This equation has only one admissible root σ$$ = S which is the ML estimator provided µ is known. If µ is not known it is replaced by y (the solution of dlnL/dµ = 0), and n s2 = ∑ (y i − y) 2 / (n − 1) i=1 σ2. 8) may be noted; we will refer to it from time to time. 9) and E(s) ≅ σ for large n. The sample standard deviation s is, therefore, fully efficient for large n.