Computation of expansions for the maximum likelihood estimator and its distribution function
In this paper, insight is given in the techniques used to compute asymptotic expansions. In a broad fashion the technique is described. Most of the results apply to the paper "An expansion for the maximum likelihood estimator and its distribution fun…
Authors: Shanti Venetiaan
COMPUT A TION OF EXP ANSIONS F OR THE MAXIMUM LIKELIHOOD ESTIMA TOR AND ITS DISTRIBUTION FUNCTION SHANTI V ENETIAAN ∗ Abstract. In this paper, insight is given in the tec hniques use d to compute asymptotic e xpansions. In a broad f ashion the te c hnique is described. Most of t he results apply to t he pa per ” An expansion for the maximum likelihoo d estimator and its distri bution function”, which will b e submitted. Key words. asymptotic expansion, maximum likelihoo d estimator of lo cation AMS sub ject classifications. 62E20 1. Expansions. In the pape r ” An expansio n for the maximum likelihoo d estimato r and its distribution function”, which will b e submitted, many expansions are being calcula ted. Because the technique itself is no t so difficult a nd the outcomes tak e a lot of space to present, many steps of the computation of the expansions will be omitted in that pap er . How ever if one wan ts to chec k the computations, it may b e useful to have some insight in how the ex pansions were obtained. None of the applied techniques are cla imed by the author. They a re just wr itten down. W e will give the full outcome of all the steps needed to construct the needed expa nsions. W e will not g ive a ny pro ofs, just the metho d of obtaining the expansions . This means that certain rest terms will be omitted. Also note that the results are only v alid for the maximum lieliho o d estimato r of lo cation. 2. An e xpansion for the m axim um lik eliho o d estimator. W e define the ma ximum likelihoo d estimator ˆ θ n by L n ( ˆ θ n ) = inf θ ∈ R L n ( θ ) , (2.1) where L n ( θ ) = n − 1 P n i =1 ρ ( X i − θ ), where ρ ( · ) = − log( · ). It may b e prov ed that (cf. Chibisov (1973)) ˆ θ n is the solutio n of the equa tion L ′ n ( θ ) = 0 , (2.2) with probability 1 + o ( n − 3 / 2 ). W e expand L ′ n ( θ ) with a T aylor expansion and get L ′ n ( θ ) = 1 n n X i =1 ρ ′ ( X i − θ ) = 1 n n X i =1 ρ ′ ( X i ) − θ n n X i =1 ρ ′′ ( X i ) + θ 2 2 n n X i =1 ρ (3) ( X i ) (2.3) − θ 3 6 n n X i =1 ρ (4) ( X i ) + θ 4 24 n n X i =1 ρ (5) ( X i − θ ′ ) , where | θ ′ | ≤ | θ | . ∗ F acult y of T ec hnology , An ton de Kom Universiteit v an Suriname, Paramarib o, Suri name 1 2 SHANTI VENETIAAN W e intro duce the notation ξ j = 1 √ n n X i =1 ( ρ ( j ) ( X i ) − a j ) , a j = E 0 ρ ( j ) ( X ) for j = 1 , ..., 5 . (2.4) Note that the ξ ′ j s are normalized sums and that a 1 = 0. (2.3) be c o mes L ′ n ( θ ) = ξ 1 n √ n − θ ( ξ 2 n √ n + a 2 ) + θ 2 2 ( ξ 3 n √ n + a 3 ) − θ 3 6 ( ξ 4 n √ n + a 4 ) + θ 4 24 ( ξ 5 n √ n + a 5 ) + · · · , (2.5) T o find the ex pansion for ˆ θ n we put ˆ θ n = B 1 / √ n + B 2 /n + B 3 /n 3 / 2 + B 4 /n 2 . (2.6) Substituting this into (2.5) leads to ( − a 2 B 1 + ξ 1 ) /n 1 / 2 + ( − ξ 2 B 1 − a 2 B 2 + 1 2 a 3 B 2 1 ) /n + ( − ξ 2 B 2 − 1 6 a 4 B 3 1 + 1 2 ξ 3 B 2 1 + a 3 B 1 B 2 − a 2 B 3 ) /n 3 / 2 (2.7) + ( − 1 2 a 4 B 2 1 B 2 + 1 2 a 3 B 2 2 + 1 24 a 5 B 4 1 − ξ 2 B 3 + ξ 3 B 1 B 2 − 1 6 ξ 4 B 3 1 + a 3 B 1 B 3 − a 2 B 4 ) /n 2 Now we tak e the first term of (2.7) and put it equal to 0. W e g et − a 2 B 1 + ξ 1 = 0 ⇒ B 1 = ξ 1 /a 2 . (2.8) Substituting the obtained B 1 in (2.7) gives ( − ξ 2 ξ 1 /a 2 + 1 2 a 3 ξ 2 1 /a 2 2 − a 2 B 2 ) /n (2.9) + ( − a 2 B 3 + a 3 ξ 1 B 2 /a 2 − ξ 2 B 2 + 1 2 ξ 3 ξ 2 1 /a 2 2 − 1 6 a 4 ξ 3 1 /a 3 2 ) /n 3 / 2 + ( a 3 ξ 1 B 3 /a 2 + 1 24 a 5 ξ 4 1 /a 4 2 − 1 2 a 4 ξ 2 1 B 2 /a 2 2 − 1 6 ξ 4 ξ 3 1 /a 3 2 + 1 2 a 3 B 2 2 + ξ 3 ξ 1 B 2 /a 2 − ξ 2 B 3 − a 2 B 4 ) /n 2 Note that the 1 / √ n ter m has v anished. W e tak e the first term o f (2.9) and put it equa l to 0. This results in B 2 = − ξ 1 ξ 2 /a 2 2 + 1 2 ξ 2 1 a 3 /a 3 2 . (2.10) W e substitute this B 2 in (2.9) and get ( − a 2 B 3 − 1 6 a 4 ξ 3 1 /a 3 2 + 1 2 ξ 3 ξ 2 1 /a 2 2 − 3 2 ξ 2 ξ 2 1 a 3 /a 3 2 + ξ 2 2 ξ 1 /a 2 2 + 1 2 ξ 3 1 a 2 3 /a 4 2 ) /n 3 / 2 (2.11) + ( − 1 6 ξ 4 ξ 3 1 /a 3 2 + a 3 ξ 1 B 3 /a 2 + 1 8 ξ 4 1 a 3 3 /a 6 2 + 1 2 ξ 3 ξ 3 1 a 3 /a 4 2 − 1 2 ξ 3 1 ξ 2 a 2 3 /a 5 2 − ξ 2 B 3 − ξ 3 ξ 2 1 ξ 2 /a 3 2 + 1 24 a 5 ξ 4 1 /a 4 2 + 1 2 a 3 ξ 2 1 ξ 2 2 /a 4 2 + 1 2 a 4 ξ 3 1 ξ 2 /a 4 2 − a 2 B 4 − 1 4 a 4 ξ 4 1 a 3 /a 5 2 ) /n 2 CALCULA TION OF EXP AN SIONS F OR THE MLE AN D ITS DISTRIBUTION FUN CTION 3 Again we put the first term of the result to 0 and get B 3 = ξ 1 ξ 2 2 /a 3 2 − 1 6 ξ 3 1 a 4 /a 4 2 + 1 2 ξ 2 1 ξ 3 /a 3 2 − 3 2 ξ 2 1 ξ 2 a 3 /a 4 2 + 1 2 ξ 3 1 a 2 3 /a 5 2 . (2.12) W e substitute B 3 in (2.11) to obtain − a 2 B 4 − 5 2 ξ 3 1 ξ 2 a 2 3 /a 5 2 + 5 8 ξ 4 1 a 3 3 /a 6 2 − ξ 3 2 ξ 1 /a 3 2 + ξ 3 ξ 3 1 a 3 /a 4 2 − 3 2 ξ 3 ξ 2 1 ξ 2 /a 3 2 (2.13) + 2 3 a 4 ξ 3 1 ξ 2 /a 4 2 + 1 24 a 5 ξ 4 1 /a 4 2 − 1 6 ξ 4 ξ 3 1 /a 3 2 + 3 a 3 ξ 2 1 ξ 2 2 /a 4 2 − 5 12 a 4 ξ 4 1 a 3 /a 5 2 . This is put equal to 0 a nd at last we obta in B 4 = ξ 3 1 ξ 3 a 3 /a 5 2 − 5 2 ξ 3 1 ξ 2 a 2 3 /a 6 2 + 5 8 ξ 4 1 a 3 3 /a 7 2 − ξ 1 ξ 3 2 /a 4 2 − 1 6 ξ 3 1 ξ 4 /a 4 2 (2.14) − 3 2 ξ 2 1 ξ 3 ξ 2 /a 4 2 + 2 3 ξ 3 1 a 4 ξ 2 /a 5 2 + 1 24 ξ 4 1 a 5 /a 5 2 + 3 ξ 2 1 a 3 ξ 2 2 /a 5 2 − 5 12 ξ 4 1 a 4 a 3 /a 6 2 . Even tua lly √ n ( ˆ θ n ) = ξ 1 a 2 + 1 √ n ( − ξ 1 ξ 2 a 2 2 + a 3 ξ 2 1 2 a 3 2 ) + 1 n ( ξ 1 ξ 2 2 a 3 2 − 3 a 3 ξ 2 1 ξ 2 2 a 4 2 + ξ 2 1 ξ 3 2 a 3 2 + a 2 3 ξ 3 1 2 a 5 2 − a 4 ξ 3 1 6 a 4 2 ) + 1 n 3 / 2 3 ξ 2 1 a 3 ξ 2 2 a 5 2 + 5 ξ 4 1 a 3 3 8 a 7 2 − 5 ξ 4 1 a 4 a 3 12 a 6 2 − 3 ξ 2 1 ξ 3 ξ 2 2 a 4 2 − 5 ξ 3 1 a 2 3 ξ 2 2 a 6 2 + ξ 4 1 a 5 24 a 5 2 (2.15) + ξ 3 1 ξ 3 a 3 a 5 2 + 2 ξ 3 1 a 4 ξ 2 3 a 5 2 − ξ 3 1 ξ 4 6 a 4 2 − ξ 1 ξ 3 2 a 4 2 + · · · , 3. Expansion for the di stribution function of the maxim um likeliho o d estima- tor. The estimator in (2.15) fits the mo del of Hall (1992 ), Section 2.3. This mea ns that the cum ulants of √ n ˆ θ n will determine the expansion for the distribution function. Fir s t we note that the cumulan ts will be o f the form (cf. Hall(19 92), Section 2.3) κ 1 = 0 + k 12 / √ n + k 13 /n 3 / 2 + · · · κ 2 = k 21 + k 22 /n + · · · κ 3 = k 31 / √ n + k 32 /n 3 / 2 + · · · (3.1) κ 4 = k 41 /n + · · · κ 5 = k 51 /n 3 / 2 + · · · . F urthermo re (cf. Hall(199 2), Section 2.2) κ 1 = E S n κ 2 = E S 2 n − ( E S n ) 2 κ 3 = E ( S n − E S n ) 3 = E S 3 n − 3 E S 2 n E S n + 2 ( E S n ) 3 (3.2) κ 4 = E ( S n − E S n ) 4 − 3 κ 2 2 κ 5 = E ( S n − E S n ) 5 − 1 0 κ 2 κ 3 . 4 SHANTI VENETIAAN 3.1. Computation o f the k appa’s. W e will now compute the k appa’s. F or the compu- tation of the k appa ’s w e need to calculate the exp ectation o f the normalized s ums, the so ca lled ξ j ’s. T erms that be c o me to o small will b e omitted. First we will in tr o duce the notation ψ i ( · ) = f ( i ) f ( · ) (3.3) η 2 = E ( ψ 2 2 ( X i )) , η 3 = E ( ψ 3 1 ( X i )) , η 4 = E ( ψ 4 1 ( X i )) , (3.4) η 5 = E ( ψ 5 1 ( X i )) , η 6 = E ( ψ 2 ψ 3 ( X i )) the ab ov e r esults in a 1 = 0 , a 2 = E ( ψ 2 1 ( X i )) , without loss of gener ality we put a 2 = 1 , (3.5) a 3 = − 1 2 η 3 , a 4 = 2 3 η 4 − η 2 , a 5 = 5 η 6 − 3 2 η 5 . and that E ( ψ 1 ψ 2 ) = 1 2 η 3 , E ( ψ 1 ψ 3 ) = − η 2 + 2 3 η 4 , E ( ψ 1 ψ 4 ) = − 5 η 6 + 3 2 η 5 E ( ψ 2 1 ψ 2 ) = 2 3 η 4 , E ( ψ 1 ψ 2 2 ) = 2 η 6 , E ( ψ 3 1 ψ 2 ) = 3 4 η 5 , E ( ψ 2 1 ψ 3 ) = − 4 η 6 + 3 2 η 5 F urthermo re w j = − ( ρ ( j ) ( X i ) − a j ) , for j = 1 , ..., 5 . Consequently , w 1 = ψ 1 , (3.6) w 2 = ψ 2 − ψ 2 1 + 1 , w 3 = ψ 3 − 3 ψ 1 ψ 2 + 2 ψ 3 1 + a 3 , w 4 = ψ 4 − 4 ψ 1 ψ 3 + 1 2 ψ 2 1 ψ 2 − 3 ψ 2 2 − 6 ψ 4 1 + a 4 w 5 = ψ 5 − 5 ψ 1 ψ 4 + 2 0 ψ 2 1 ψ 3 − 1 0 ψ 2 ψ 3 + 3 0 ψ 1 ψ 2 2 − 6 0 ψ 3 1 ψ 2 + 2 4 ψ 5 1 + a 5 Note that E w j = 0 fo r j = 1 , · · · 5 and that E ( w 2 1 ) = 1. E ( w 1 w 2 ) = − 1 2 η 3 , E ( w 1 w 3 ) = 2 3 η 4 − η 2 , E ( w 1 w 4 ) = 5 η 6 − 3 2 η 5 E ( w 2 1 w 2 ) = − 1 3 η 4 + 1 , E ( w 2 1 w 3 ) = − 4 η 6 + 5 4 η 5 − 1 2 η 3 E ( w 3 1 w 2 ) = − 1 4 η 5 + η 3 , E ( w 1 w 2 2 ) = 2 η 6 − 1 2 η 5 − η 3 CALCULA TION OF EXP AN SIONS F OR THE MLE AN D ITS DISTRIBUTION FUN CTION 5 E ( w 2 2 ) = η 2 − 1 3 η 4 − 1 , E ( w 2 w 3 ) = − η 6 + 1 4 η 5 + 1 2 η 3 W e will give an example here to illustra te how the expectatio ns of the terms of √ n ˆ θ n are obtained. E ξ 2 1 = E ( 1 √ n X i w 1 i ) 2 = E { 1 n ( X i w 2 1 i + 2 X X i
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