Further, using the definition of g$_{j}$ ( z$_{ρ}$$_{(}$$_{j}$$_{)0}$ ) , the fact that ∑ $_{ρ}$sgn ρz j − 1 ρ ( j )0 equals the Vandermonde determinant ∏ r i>j ( z$_{i}$$_{0}$ − z$_{j}$$_{0}$ ) , we get Substituting (52) and (53) into (51), and using (49) and (50) together with the fact that the branch of the square root in f − 1 / 2 i 2 is chosen so that √ − 1 = − i , we get after some algebra Now, using the fact that S − p = O$_{p}$ (1) , we get ln ( S p ) = S − p p + O$_{p}$ ( p − $^{2}$) and Further, the Stirling approximations give 32 ( p − t )! = e − $^{p}$pp − t +1 ( 2 π p ) 1 / 2 ( 1 + O ( p − $^{1}$)) and Γ ( p ( n − r ) + r ( r + 1) / 2) Γ ( np ) = ( pn ) − pr + r ( r +1) / $^{2}$e 1 $_{2}$c$_{p}$ r 2 ( 1 + O ( n − . ln ( 1 − r ∑ j =1 h$_{j}$ + c$_{p}$ S ) = − ∑ r j $_{=1}$( h$_{j}$ + c$_{p}$ ) p − 1 $_{2}$( ∑ r j $_{=1}$( h$_{j}$ + c$_{p}$ ) 2 p 2 + ∑ r j $_{=1}$( h$_{j}$ + c$_{p}$ ) p 2 ( S − p ) + O$_{p}$ ( p − $^{3}$) . L ( h ; µ ) = r ∏ t =1 [ ( p − t )! ( c$_{p}$ − h 2 t ) 1 / 2 ] S pr − r ( r +1) / 2 × Γ ( p ( n − r ) + r ( r + 1) / 2) Γ ( np ) ( 1 − r ∑ j =1 h$_{j}$ + c$_{p}$ S ) − p ( n − r ) − r ( r +1) / 2 × exp { − r ∑ j =1 nh$_{j}$ } 1 (2 πn ) r/ $_{2}$c − pr p r ∏ i>j ( c$_{p}$ − h$_{i}$h$_{j}$ ) ( 1 + O$_{p}$ ( n − $^{1}$)) . ∑ ρ sgn ρ r ∏ j =1 e − $^{nf$_{ρ}$$_{(}$$_{j}$$_{)0}$}$g$_{j}$ ( z$_{ρ}$$_{(}$$_{j}$$_{)0}$ ) π 1 / 2 f 1 / 2 ρ ( j $_{)2}$n 1 / 2 = π r/ 2 n r/ 2 exp { − r ∑ i =1 ( nf$_{i}$$_{0}$ + Δ$_{p}$ ( z$_{i}$$_{0}$ )) } × (53) r ∏ i =1 f − 1 / 2 i 2 r ∏ i>j ( z$_{i}$$_{0}$ − z$_{j}$$_{0}$ ) .