5.2 Proof of Theorem 3 Setting θ$_{pj}$ = 1 2 c$_{p}$ h$_{j}$ 1+ $_{h$_{j}$}$, we have v$_{pj}$ = 1 + h$_{j}$ , θ$_{pj}$v$_{pj}$ = h$_{j}$ 2 $_{c$_{p}$}$, and Turning to the proof of (11), consider the integrals Further, by Lemma 11 and formula (3.3) of OMH, ∫ ln ( z$_{j}$$_{0}$ − λ ) d F MP p ( λ ) = h$_{j}$ c$_{p}$ − 1 c$_{p}$ ln (1 + h$_{j}$ ) + ln (1+ h$_{j}$ ) c$_{p}$ h$_{j}$ for sufficiently large p, a.s.. With these auxiliary results, formula (10) is a straightforward consequence of (3) and Proposition 2. In what follows, we omit the subscript p in n$_{p}$ to simplify notation. Note that I (0 , ∞ ) is the integral appearing in expression (4) for L$_{p}$ ( h ; µ$_{p}$ ) . Let us now prove that, for some constant α > 0 , where o (1) is uniform in h ∈ [0 $_{,}$√$_{c}$ − δ ] $^{r}$. Since, by Corollary 1, S$_{p}$/p → 1 a.s., the set H$_{δ}$ is bounded from below, and λ$_{p}$$_{1}$ → (1 + √$_{c}$$_{)}$ 2 a.s., there exists a constant A$_{1}$ > 0 that depends only on δ and r, such that inf [ 0 $_{,}$√$_{c}$ − δ ] r px tr ( D$_{p}$Q ′ Λ$_{p}$ Q ) /S$_{p}$ ≥ − A$_{1}$x/ 2 for all x ≥ 0 and all sufficiently large p, a.s.. Therefore, for all h ∈ [0 $_{,}$√$_{c}$ − δ ] $^{r}$, and, using Stirling’s approximation, we get Next, there exists a constant A$_{2}$ > 0 such that, for all x ≥ 0 and all sufficiently large p, $^{sup$_{h}$}$∈ [ 0 $_{,}$√$_{c}$ − δ ] r px tr ( D$_{p}$Q ′ Λ$_{p}$ Q ) /S$_{p}$ ≤ A$_{2}$x/ 2, a.s.. Therefore, a.s., for all 33 I (0 , ∞ ) ≥ ( n + A$_{1}$ 2 ) − np 2 ( np 2 ) np 2 $_{e}$− np 2 ( 4 π np ) 1 / 2 (1 + o (1)) = p np 2 $_{$_{e}$−}$( n $_{2}$+ A 1 2 − 1 4 A 2 1 n ) p ( 4 π np ) 1 / 2 (1 + o (1)) , a.s. (51) 2 I (0 , ∞ ) ≥ ∫ ∞ 0 x np 2 − 1 $_{e}$− n + A 1 2 $^{x}$d x = ( n + A$_{1}$ 2 ) − np 2 Γ ( np 2 ) , I (0 , ∞ )= I ( p − α √ p, p + α √ p ) (1+ o (1)) , a.s. (50) I ( k$_{1}$, k$_{2}$ )= ∫ k$_{2}$ k$_{1}$ x n p p 2 − 1 $_{e}$− n p 2 x ∫ O ( p ) e p x S p tr( D$_{p}$Q ′ Λ$_{p}$ Q $^{)}$(d Q )d x, k$_{1}$ < k$_{2}$ ∈ R . ln (1+2 θ$_{pj}$v$_{pj}$ − 2 θ$_{pj}$λ$_{pi}$ ) = ln ( 1 c$_{p}$ h$_{j}$ 1 + h$_{j}$ ) + ln ( z$_{0}$$_{j}$ − λ$_{pi}$ ) .