Page 291 - Elements of Distribution Theory

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

9.5 Laplace’s Method 277

so that u is a one-to-one function of y with
1 2
u = g(ˆ y) − g(y).
2
Then

udu =−g (y) dy.
Note that, since g is maximized at ˆ y and |g (y)| > 0 for all y = ˆ y,it follows that g (y) < 0

for y > ˆ y and g (y) > 0 for y < ˆ y. Hence, u and g (y)have the same sign and

u

dy
= .
du g (y)

It follows that
b u(b) n u
h(y)exp{−n[g(ˆ y) − g(y)]} dy = h(y(u)) exp − u 2 du

a u(a) 2 g (y(u))
u(b) n
¯
≡ h(u)exp − u 2 du.
u(a) 2
Under the conditions on h and g,
¯
2
¯
¯
h(u) = h(0) + h (0)u + O(u )as u → 0
and, hence, by Corollary 9.1,
√
1
u(b) n 2
1
¯
h(u)exp − u 2 du = √ 2 ¯ h(0) + O 3 as n →∞.
u(a) 2 n n 2
Hence, to complete the approximation we need to find
uh(y(u))
¯
h(0) ≡ lim .
u→0 g (y(u))

Note that u = 0if and only if y = ˆ y; hence,
h(y(0)) = h(ˆ y), g (y(0)) = 0

and
u
¯
h(0) = h(ˆ y) lim ≡ h(ˆ y)L
u→0 g (y(u))

where
u
L = lim .
u→0 g (y(u))

By L’Hospital’s rule,
1
L = .
g (ˆ y)y (0)

We have seen that
u

y (u) =
−g (y(u))