Sequence Comparison.pdf

7.2 Ungapped Local Alignment Scores 129
liminf (1 −
x→∞
F(x))eλx ≥ liminf (1 −
x→∞ F(⌈x⌉))eλ⌈x⌉ = C
e λ − 1 = C′ .
On the other hand, because limsup x→∞ x −⌊x⌋ = 1,
limsup (1 − F(x))e λx ≤ lim (1 − F(⌊x⌋))e ⌊x⌋ e λ(x−⌊x⌋) ≤
C
x→∞
x→∞ e λ − 1 eλ = C ′ e λ .
Replacing x by ln(m)/λ + y, (7.29) becomes
Pr[M(K m ) ≤ ln(m)/λ + y]=(F(ln(m)/λ + y)) m =
1
exp{−mln(F(ln(m)/λ + y))} .
lnz
Using lim z→1 z−1
= 1, we obtain
limsup Pr[M(K m ) ≤ ln(m)/λ + y]
m→∞
1
= limsup
m→∞ exp{−mln(F(ln(m)/λ + y))}
1
=
liminf m→∞ exp{m(1 − F(ln(m)/λ + y))}
1
≤
liminf m→∞ exp { mC ′ e −(ln(m)+λy)}
{
= exp −C ′ e −λy} .
Similarly,
liminf Pr[M(K m) ≤ ln(m)/λ + y]
m→∞
1
= liminf
m→∞ exp{−mln(F(ln(m)/λ + y))}
1
=
limsup m→∞ exp{m(1 − F(ln(m)/λ + y))}
1
≥
limsup m→∞ exp { mC ′ e λ e −(ln(m)+λy)}
{
= exp −C ′ e λ e −λy} .
Hence, the lemma is proved.
⊓⊔
Now consider the accumulative score walk W corresponding to an ungapped
alignment of length n.Let