sufficiently large p , where Γ ( np 2 , y ) is the complementary incomplete Gamma function (see Olver 1997, p.45) with y = ( p + α √ p ) ( n − A$_{2}$ 2 ) . Hence, for sufficiently large p, y > np/ 2 + nα √ p/ 4 , and we can continue Now, Γ ( β, γ ) ≤ e − $^{γ}$γ$^{β}$/ ( γ − β + 1) whenever β > 1 and γ > β − 1 (Olver 1997, p.70). Therefore, we have, for sufficiently large p , Comparing this to (51), we see that α can be chosen so that Further, for sufficiently large p , where y = ( p − α √ p ) n − A$_{2}$ 2 < np 2 − αn √ p 4 . Therefore, for any positive z < np 2 and sufficiently large p, 34 I (0 , p − α √ p ) ≤ ( n − A$_{2}$ 2 ) − np 2 ∫ np 2 − αn √ p 4 0 t np 2 − 1 $_{e}$− $^{t}$d t < n − A$_{2}$ 2 − np 2 ( np 2 − αn √ p 4 ) z Γ ( np 2 − z ) . I (0 , p − α √ p ) ≤ ∫ p − α √ p 0 x np 2 − 1 $_{e}$− $^{n}$−A 2 2 $^{x}$d x = ( n − A$_{2}$ 2 ) − np 2 ∫ y 0 t np 2 − 1 $_{e}$− $^{t}$d t, a.s., I ( p + α √ p, ∞ ) = o (1) I (0 , ∞ ) , a.s. (52) I ( p + α √ p, ∞ ) < ( 1 − A$_{2}$ n ) − np 2 e − np 2 − αn √ p 4 p np 2 ( 1 + α 2 √ p ) np 2 αn √ p/ 4 + 1 = p np 2 e A 2 p + A 2 p 4 n $_{e}$− np 2 − α 2 n 16 + α 3 n 48 √ p − α 4 n 128 p αn √ p/ 4 + 1 (1 + o (1)) < p np 2 $_{e}$− np 2 e p ( A$_{2}$ − α 2 n 32 p ) αn √ p/ 4 + 1 (1 + o (1)) , a.s. I ( p + α √ p, ∞ ) < ( n − A$_{2}$ 2 ) − np 2 Γ ( np 2 , np 2 + αn √ p 4 ) , a.s. I ( p + α √ p, ∞ ) ≤ ∫ ∞ p + α √ p x np 2 − 1 $_{e}$− $^{n}$−A 2 2 $^{x}$d x = ( n − A$_{2}$ 2 ) − np 2 Γ ( np 2 , y ) ,