Mathematical Methods for Physics and Engineering - Matematica.NET

31.4 SOME BASIC ESTIMATORSwhere s 4 is given bys 4 =[ ∑i x2 iN( ∑− i x ) ] 2 2iN= (∑ i x2 i )2− 2 (∑ i x2 i )(∑ i x i) 2+ (∑ i x i) 4. (31.45)N 2 N 3 N 4We will consider in turn each of the three terms on the RHS. In the first term, the sum( ∑ i x2 i )2 can be written as( ) 2∑x 2 i = ∑ x 4 i + ∑ x 2 i x 2 j ,iii,jj≠iwhere the first sum contains N terms and the second contains N(N − 1) terms. Since thesample elements x i and x j are assumed independent, we have E[x 2 i x2 j ]=E[x2 i ]E[x2 j ]=µ2 2 ,and so⎡( ) ⎤ 2∑E ⎣ x 2 ⎦i = Nµ 4 + N(N − 1)µ 2 2.iTurning to the second term on the RHS of (31.45),( )( ) 2∑ ∑x 2 i x i = ∑ x 4 i + ∑ x 3 i x j + ∑ x 2 i x 2 j + ∑ x 2 i x j x k .iiii,ji,ji,j,kj≠ij≠ik≠j≠iSince the mean of the population has been assumed to equal zero, the expectation valuesof the second and fourth sums on the RHS vanish. The first and third sums contain Nand N(N − 1) terms respectively, and so⎡( )( ) ⎤ 2∑ ∑E ⎣ x 2 i x i⎦ = Nµ 4 + N(N − 1)µ 2 2.iiFinally, we consider the third term on the RHS of (31.45), and write( ) 4∑x i = ∑ x 4 i + ∑ x 3 i x j + ∑ x 2 i x 2 j + ∑ x 2 i x j x k + ∑ x i x j x k x l .iii,ji,ji,j,ki,j,k,lj≠ij≠ik≠j≠il≠k≠j≠iThe expectation values of the second, fourth and fifth sums are zero, and the first and thirdsums contain N and 3N(N − 1) terms respectively (for the third sum, there are N(N − 1)/2ways of choosing i and j, and the multinomial coefficient of x 2 i x 2 j is 4!/(2!2!) = 6). Thus⎡( ) ⎤ 4∑E ⎣ x i⎦ = Nµ 4 +3N(N − 1)µ 2 2.iCollecting together terms, we therefore obtainE[s 4 (N − 1)2]= µN 3 4 + (N − 1)(N2 − 2N +3)µ 2N2, (31.46)3which, together with the result (31.43), may be substituted into (31.44) to obtain finallyV [s 2 (N − 1)2 (N − 1)(N − 3)]= µN 3 4 − µ 2N 3 2= N − 1 [(N − 1)νN 3 4 − (N − 3)ν2], 2 (31.47)1247