Normal approximation to the hypergeometric distribution in ...

S.N. Lahiri et al. / Journal of Statistical Planning and Inference 137 (2007) 3570 –3590 3587
Hence, for −1x 0, noting that K0 − K1 ,
|1(x)|
K0
j=K1
exp(−˜x 2 j (0.07)) exp(−1 ) 5a1
√ 2 2
(K0 − K1) exp( −1 ) 5a1
√ 2 2
C
. (4.29)
Thus, the bound (4.28) on I2 holds for all x ∈[−, 0].
Next consider I1. Note that for j ∈{0, 1,...,n},
P(X= j + 1)P(X= j)
Np − j
⇔
j + 1 .
n − j
Nq − n + j + 1 1
n(Np + 1) Nq + 1
⇔ j − . (4.30)
N + 2 N + 2
Thus, P(X= j)