Page 286 - Elements of Distribution Theory

P1: JZP
052184472Xc09 CUNY148/Severini June 2, 2005 12:8

272 Approximation of Integrals

Consider the integral

T n
I n = h n (t)exp{−nt} dt
0
where T 1 , T 2 ,... are bounded away from 0. Then
m
c j (a j + 1) 1
I n = + O .
a
n a j +1 n m+1 +1
j=0
Proof. Fix > 0 such that ≤ T n for all n. There exists a constant K such that for all
n = 1, 2,...,
|h n (t)|≤ K exp{bt}, t ≥ .
Hence, for sufficiently large n,

T n ∞ K
h n (t)exp{−nt} dt ≤ K exp{−(n − b)t} dt = exp{−(n − b) }

n − b

exp{−n }
= O as n →∞.
n
Consider the integral

ˆ
h n (t)exp{−nt} dt = h nm (t)exp{−nt} dt + R nm (t)exp{−nt} dt.
0 0 0
For any > 0,
m
c nj (a j + 1)
ˆ
h nm (t)exp{−nt} dt = a j +1 + O(exp{−n })
0 j=0 n
and
(a m+1 + 1)

|R nm (t)| exp{−nt} dt ≤ K m a .
0 n m+1 +1
The theorem follows from combining these results.

Example 9.6 (Tail probability of the t-distribution). Consider the integral
(ν+1)
∞ 2
y
2 −
1 + dy
z ν
which, aside from a normalizing constant, is the tail probability of the t-distribution with ν
degrees of freedom. We will derive an asymptotic expansion for this integral as ν →∞.
Using the change-of-variable
1 y 2 1 z 2
t = log 1 + − log 1 + ,
2 ν 2 ν
we may write
∞ (ν+1) ν ∞ 1
√
2
(1 + y /ν) − 2 dy = ν (1 − exp(−2t)/c ν ) − 2 exp(−νt) dt
z c ν 2 0