“…in the proof of Theorem 1 and Proposition 3 we furthermore know that c je,ke;De (u, v) = c je,ke;De (u, v)+ o p {b 2 n,c + (nb 2 n,c ) −1/2 } as well as F je|De (x je |x De ) = F je|De (x je |x De ) + o p {b 2 n,c + (nb 2 n,c ) −1/2 }. Similar to Lemma B3, we can now show that c je,ke;De F je|De (x je |x De ), F ke|De (x ke |x De ) = c je,ke;De F je|De (x je |x De ), F ke|De (x ke |x De ) + o p {b 2 n,c + (nb 2 n,c ) −1/2 }.Hence, for(16) to hold it suffices to show that(nb 2 n,c ) 1/2 c * (x) − b 2 n,c μx − c * (x) d → N 0, Σx ,(25)wherec * (x) = cje,ke;De {F je|De (x je |x De ), F ke|De (x ke |x De )} e∈E 1 ,...,E d−1 , and c * (x) is defined similarly, but replacing c je,ke;De with c je,ke;De . ke|De ), z je|De := Φ −1 F je|De (x je |x De ) , z ke|De := Φ −1 F ke|De (x ke |x De ) .…”