DReynold = − ln ( 1 − θ ) DReynold = -ln (1 - %theta)

the following notation is used, for all 𝚹 estimators:

n ˉ = ∑ i = 1 r n i r , n c = ( r n ˉ − ∑ i = 1 r n i 2 r n ˉ ) ( r − 1 ) , p lu ˜ = ∑ i = 1 r n i p ilu ˜ r n ˉ , α il ˜ = 1 − ∑ u = 1 v 1 p ilu 2 ˜ bar n = sum _{i=1} ^r {{n_i} over r } , n_c = (r bar n - sum _{i=1} ^r {n^2 _i over {r bar n}}) over (r-1), tilde p _lu = sum _{i=1} ^r {n_i tilde p _ilu} over {r bar n}, tilde %alpha _il = 1 - sum ^{v_1} _{u=1} tilde p _ilu ^2

θ L ˜ = 2x + y − z ± ( z − y ) 2 + 4x 2 2 ( y − z ) tilde{%theta }_{L} = {2x+y-z +- sqrt{( z-y) ^{2}+4x^{2}}} over {2( y-z) }

where: z = ∑ l = 1 m a l 2 z= sum ^{m}_{l=1}a_{l}^{2} , x = ∑ l = 1 m a l b l x= sum ^{m}_{l=1}a_{l}b_{l} and y = ∑ l = 1 m b l 2 y= sum ^{m}_{l=1}b_{l}^{2} . to check which of the two solutions for ̃𝛉 _L provides the minimum, the residual sum of squares, R, should be calculated for each where:

R = ( 2x + y + z ) θ L 2 ˜ − 2 ( x + z ) θ L ˜ + z 1 − 2 θ L ˜ + 2 θ L 2 ˜ R={ ( 2x+y+z ) tilde{%theta }_{L}^{2}-2 ( x+z ) tilde{%theta }_{L}+z} over {1-2 tilde{%theta }_{L}+2 tilde{%theta }_{L}^{2}}