Page 292 - Elements of Distribution Theory
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P1: JZP
052184472Xc09 CUNY148/Severini June 2, 2005 12:8
278 Approximation of Integrals
so that
1
y (0) = =−L
−g (ˆ y)y (0)
and, hence, that
1
2
L = .
−g (ˆ y)
Finally, note that u and g (y(u)) have the same sign so that L ≥ 0. It follows that
1
L =
1
[−g (ˆ y)] 2
and
h(ˆ y)
¯
h(0) = 1 .
[−g (ˆ y)] 2
Hence, if |h(ˆ y)| > 0, then
√
u(b) n 2 1 h(ˆ y) 1
¯
h(u)exp − u 2 du = √ 2 1 + O 3
u(a) 2 n [−g (ˆ y)] 2 n 2
so that
√
(2π)h(ˆ y) −1
I n = exp{ng(ˆ y)} 1 [1 + O(n )] as n →∞.
[−ng (ˆ y)] 2
If h(ˆ y) = 0, then
u(b) n 1
¯
h(u)exp − u 2 du = O 3
u(a) 2 n 2
so that
1
I n = exp{ng(ˆ y)}O as n →∞.
3
n 2
Example 9.8 (Stirlings approximation). Consider approximation of the gamma function
(z + 1) for large values of z. Note that, using the change-of-variable y = t/z,wemay
write
∞ ∞
z
(z + 1) = t exp(−t) dt = z z+1 exp{−z[y − log(y)]} dy.
0 0
Hence, we may apply Theorem 9.14 with g(y) = log(y) − y, n = z, and h(y) = 1. It is
straightforward to show that the conditions of Theorem 9.14 are satisfied with ˆ y = 1,
leading to the result that
√ z+1/2 −1
(z + 1) = (2π)z exp(−z)[1 + O(z )] as z →∞.
Since (z) = (z + 1)/z, Stirling’s approximation to (z)isgiven by
√ z−1/2 −1
(z) = (2π)z exp(−z)[1 + O(z )] as z →∞.