Normal approximation to the hypergeometric distribution in ...

Note that for all k, n satisfying |ak,n|,
S.N. Lahiri et al. / Journal of Statistical Planning and Inference 137 (2007) 3570 –3590 3583
(1 − f)
Np − k Np − (np + (1 − f)npq)>np > 0
2
and
(1 − f)
Nq − (n − k)>nq > 0.
2
Hence, by the error bound in Stirling’s approximation, for all k, n with |ak,n| and 6(np ∧ nq)1,
∗ 1
12Np + 1 −
1
12(Np − k) +
1
12Nq + 1 −
1
12(Nq − (n − k)) +
1 1
−
12(N − n) + 1 12N
−
−
12k + 1
(12Np + 1)(12(Np − k)) −
12(n − k) + 1
(12Nq + 1)(12(Nq − n + k))
1
6Np(1 − )(1 − f) −
1
6Nq(1 − )(1 − f)
f
=−
6npq(1 − )(1 − f) ,
∗
1 1
f
0 + 0 +
−
12(N − n) + 1 12N 6npq(1 − )(1 − f) ,
∗∗ 1
12n −
1
12k + 1 −
1
12(n − k) + 1 0,
∗∗ 1 1
−
12n + 1 12k −
1
n
1
−
−
12(n − k) 12k(n − k) 6npq(1 − ) .
Hence, the lemma follows from (4.21) and the above inequalities.
Lemma 2. Let g : R −→ [0, ∞) be such that g is ↑ on (−∞,a)andgis↓ on (a, ∞) for some a ∈ R. Then, for any
k ∈ N, b ∈ R and h ∈ (0, ∞),
k
g(b + ih)
i=o
b+hk
where g(x0) = max{g(b + ih) : i = 0, 1,...,k}.
Proof. For ba, by monotonicity,
h
k
g(b + ih)hg(b) +
i=0
b
g(x)dx + 2hg(x0), (4.22)
b+hk
b
g(x)dx.
For b