Normal approximation to the hypergeometric distribution in ...

S.N. Lahiri et al. / Journal of Statistical Planning and Inference 137 (2007) 3570 –3590 3581
By similar arguments,
A2 = Nq(1 − f)(1 + zk,n) + 1
log(1 + zk,n)
2
=
Nq(1 − f)+ 1
zk,n +
2
z2 k,n
2
where for all n, k, satisfying |zk,n|,
|zk,n|
3
|r3n(k)|Nq(1 − f)
2 +
Nq(1 − f)(1 + zk,n) + 1
2 ·
From, (4.11), (4.12) and (4.14), we have
(zk,n − yk,n)
log R(k; n, M, N) = ∗ −
+ y2 k,n
2
log(1 − f)
2
Nq(1 − f)− 1
+ r3n(k), (4.14)
2
−
2
+ z2 k,n
2
|zk,n| 3
Np(1 − f)− 1
+ r2n(k) + r3n(k)
2
. (4.15)
3 3(1 − )
Nq(1 − f)− 1
2
= ∗ − 1
2 log(1 − f)− x2 k,nf 2(1 − f) + r4n(k), (4.16)
where for all n, k satisfying (|yk,n|∨|zk,n|),
|r4n(k)||r2n(k)|+|r3n(k)|+ 1 2 |yk,n − zk,n|+ 1 4 (y2 k,n + z2 k,n ).
Next using Stirling’s formula on the binomial term, we have
n
log p
k
k q n−k
e
= log
(n−k−n−k)
√ 1
p
√ − nq − xk,n npq + log 1 − xk,n
2npq
2
nq
√ 1
q
− np + xk,n npq + log 1 + xk,n
2
np
≡ ∗∗ − log 2npq − A3 − A4 say, (4.17)
where ∗∗ √ √
= n − k − n−k. Next write ˜yk,n = xk,n p/nq and ˜zk,n = xk,n q/np. Then, by arguments similar to (4.12)
and (4.14),
√ 1
p
A3 = nq − xk,n npq + log 1 − xk,n
2
nq
=−˜yk,n nq + 1
+
2
˜y2
k,n
nq −
2
1
+ r5n(k)
2
and
√ 1
q
A4 = np + xk,n npq + log 1 + xk,n
2
np
=˜zk,n np + 1
+
2
˜z2
k,n
np −
2
1
+ r6n(k),
2