1 Introduction

The usual saddlepoint approximations of Daniels (1954) are based on asymptotic normality of tilted means. However, if the means have finite support then the approximations can fail near the endpoints of the support where asymptotic normality of the tilted means does not hold. This paper gives new accurate methods for saddlepoint approximations near the extremes of the support for binomial, hypergeometric, one and two sample Wilcoxon statistics. We will consider only the lower extremes of the support in each case, the upper extreme having analogous results.

For a binomial statistic B, saddlepoint approximations are obtained by considering the asymptotic probabilities for a tilted variable B_{t}, where t is chosen so that EB_{t} = k. There are 3 cases, namely, k equal to 0, fixed k and k tending to infinity. When k is equal to 0, the probability that B_{t} is equal to 0 is 1, when k is fixed, B_{t} converges to a Poisson variable with parameter k as noted by Jensen(1995, section 10.4) and when k tends to infinity, B_{t} is asymptotically normal. In this last case, we obtain a relative error for the saddlepoint approximation as 1/k. This extends the known result for cn < k < n/2 where 0 < c < 1/2 in which case the saddlepoint approximation has relative error 1/n.

For a hypergeometric variable H we obtain similar results to those of the binomial variable, using an approach based on conditioning.

The one-sample Wilcoxon statistic W can be expressed as the sum of independent variables. Saddlepoint approximations are obtained by considering the asymptotic probabilities for a tilted variable W_{t}, where t is chosen so that EW_{t} = k (see, for example, Robinson (1982)). Also there are 3 cases, namely, k equal to 0, fixed k and k tending to infinity. When k is equal to 0, W_{t} is equal to 0 with probability 1, when k is fixed, a new combinatorial result is obtained and when k tends to infinity, W_{t} is asymptotically normal. In this last case, we obtain a relative error for the saddlepoint approximation as 1/\sqrt{k} which extends the known result for cn(n+1)/2 < k < n(n+1)/4 where 0 < c < 1/2 in which case the relative error is 1/n.

For the two-sample Wilcoxon rank sum statistic, U_{m,n}, conditioning is used to obtain results like those of the one-sample Wilcoxon statistic.

In the sequel, for two sequences a_{k} and b_{k}, a_{k} ∼ b_{k} means there exist constants C > c > 0 such that cb_{k} < a_{k} < Cb_{k}.

2 Binomial Distribution

Assume X_{1}, …, X_{n} are i.i.d. and

P (X_{i} = 1) = p, P (X_{i} = 0) = 1 − p = q, i = 1, …, n.

Put

B = nX̄ = X_{1} + … + X_{n}.

Then define tilted random variable X_{1t},…,X_{nt} as

f_{t}(x) = P (X_{it} = x) = \frac{e^{tx}p^{x}q^{1−x}}{q + pe^{tx}} = e^{−K(t)+tx}f(x),  (2.1)

where f(x) = p^{x}q^{1−x}, x = 0, 1 and

K(t) = q + pe^{tx}.

Put

B_{t} = X_{1t} + … + X_{nt}.

Then, for x = k/n, we have

P (B = nx) = e^{−nΛ(x)}P (B_{t(x)} = nx),

where

t(x) = log \frac{(1 − p)x}{p(1 − x)}

is the solution of K′(t) = x and

Λ(x) = −K(t(x)) + xt(x) = (1 − x) log \frac{1 − x}{1 − p} + x log \frac{x}{p}.

Also,

B_{t(x)} ∼ bin(n, x)  (2.2)

because

P (X_{it} = 1) = \frac{pe^{t(x)}}{q + pe^{t(x)}} = \frac{\frac{(1−p)x}{p(1−x)}p}{q + p\frac{(1−p)x}{p(1−x)}} = x.

Consider the following 3 cases.

(a) k → ∞ and n − k → ∞

(b) k = 0(or n),

(c) k is fixed and finite integer (or n − k is fixed and finite integer). In each of these cases we have a different limiting distribution for B_{t(x)}.

Theorem 2.1 For B a binomial (n, p) variable, as n → ∞, we have (a) for k → ∞, k ≤ np,

P (B = k) = \frac{e^{−nΛ(k/n)}}{\sqrt{2πk(1 − k/n)}}(1 + O(\frac{1}{k}));

(b) for k = 0,

P (B = 0) = e^{−nΛ(0)} = q^{n};

(c) for k a fixed integer,

P (B = k) = e^{−nΛ(k/n)} \frac{k^{k}e^{−k}}{k!} (1 + O(\frac{1}{n})).  (2.3)

Proof: (a) when k → ∞, we have

\{ \begin{matrix}K′′(t) = \frac{pqe^{t}}{(q+pe^{t})^{2}} ∼ \frac{k}{n}, \\ K^{(3)}(t) = \frac{pqe^{t}(q−pe^{t})}{(q+pe^{t})^{3}} ∼ \frac{k}{n}, \\ K^{(4)}(t) = \frac{pqe^{t}[(q−2pe^{t})(q+pe^{t})−3pe^{t}(q−pe^{t})]}{(q+pe^{t})^{4}} ∼ \frac{k}{n}, \\ K^{(5)}(t) ∼ \frac{k}{n} \end{matrix}  (2.4)

So nK′′(t) → ∞ as k → ∞. We know

P (B_{t} = k) = \frac{1}{2π} ∫_{−π}^{π} exp\{−iξk + nK(t + iξ) − nK(t)\}dξ = J_{1} + J_{2},

where

J_{1} = \frac{1}{2π} ∫_{−ϵ}^{ϵ} exp\{−iξk + nK(t + iξ) − nK(t)\}dξ

and

J_{2} = \frac{1}{2π} ∫_{(−π,π)\(−ϵ,ϵ)} exp\{−iξk + nK(t + iξ) − nK(t)\}dξ.

Now we look at J_{1} first.

J_{1} = \frac{1}{2π} ∫_{−ϵ}^{ϵ} exp\{−\frac{1}{2}nK′′(t)ξ^{2} + \frac{nK^{(3)}(t)}{3!}(iξ)^{3} + \frac{nK^{(4)}(t)}{4!}(iξ)^{4} + \frac{nK^{(5)}(t + θiξ)}{5!}(iξ)^{5}\}dξ.

Put η = \sqrt{nK′′(t)}ξ. Then, from

\frac{nK^{(3)}(t)}{(nK′′(t))^{3/2}} ∼ \frac{1}{k^{1/2}} , \frac{nK^{(4)}(t)}{(nK′′(t))^{2}} ∼ \frac{1}{k}, \frac{K^{(5)}(t)}{(K′′(t))^{5/2}} ∼ \frac{1}{k^{3/2}} ,

when ϵ > 0 is small enough, we have

J_{1} = \frac{1}{2π\sqrt{nK′′(t)}} ∫_{−ϵ\sqrt{nK′′(t)}}^{ϵ\sqrt{nK′′(t)}} e^{− \frac{1}{2} η^{2}}(1 + \frac{K^{(3)}(t)}{3!K′′(t)^{3/2}\sqrt{n}}(iη)^{3} + \frac{K^{(4)}(t)}{4!K′′(t)^{2}n}(iη)^{4}

+\frac{K^{(3)}(t)^{2}}{3!K′′(t)^{3}n}(iη)^{6} + P_{5}(η)e^{\frac{1}{4} η^{2}})dη,

where P_{5}(η) is the polynomial of degree 5. Because

\frac{K^{(3)}(t)}{(K′′(t))^{3/2}\sqrt{n}} ∼ \frac{1}{k^{1/2}} , \frac{K^{(4)}(t)}{(K′′(t))^{2}n} ∼ \frac{1}{k},

we have

J_{1} = \frac{1}{\sqrt{2πnK′′(t)}}(1 + O(\frac{1}{k})).

Now let us look at J_{2}. J_{2} is rewritten as

J_{2} = \frac{1}{2π} ∫_{(−π,π)\(−ϵ,ϵ)} e^{−iξk} (\frac{q + pe^{t+iξ}}{q + pe^{t}} )^{n} dξ.

When ϵ ≤ |ξ| ≤ π, we have

|\frac{q + pe^{t+iξ}}{q + pe^{t}} |^{n} ≤ \frac{((q + pe^{t})^{2} − 2pqe^{t}(1 − cosϵ))^{n/2}}{(q + pe^{t})^{n}} = (1 − 2(1 − \frac{k}{n})(1 − cosϵ)\frac{k}{n})^{n/2} .

We know x = k/n ≤ p. So, when n is big enough and ϵ < π/3 , we have

|\frac{q + pe^{t+iξ}}{q + pe^{t}} |^{n} ≤ (1 − \frac{δk}{n})^{n/2} = e^{−δ_{1}k/2},

where 0 < δ, δ_{1} < 1. Therefore,

J_{2} = O(k^{−3/2}).

We know

P (B = k) = e^{−nΛ(x)}P (S_{t} = k).

Therefore,

P (B = nx = k) = \frac{e^{−nΛ(x)}}{\sqrt{2πnx(1 − x)}}(1 + O(\frac{1}{k})).

(b) For k = 0, from (2.2), we have

B_{t(x)} = B_{t(0)} ∼ bin(n, 0).

So P (B_{t(0)} = 0) = 1 and the correct saddlepoint approximation is

P (B = 0) = e^{−nΛ(0)}1 = (1 − p)^{n} = q^{n}.

This is indeed the exact probability.

(c) The tilted variable B_{t(x)} is binomial (n, x). Since nx = k is fixed, B_{t(x)} converges in distribution to a Poisson(k) variable. Indeed

P (B_{t(x)} = k) = \frac{k^{k}e^{−k}}{k!} (1 + O(\frac{1}{n}))

and (2.3) follows. □

Remark: In (b), The usual saddlepoint approximation tends to ∞ as n → ∞. In (c), we can use Stirlings formula to obtain

P (B = k) = e^{−nΛ(k/n)} \frac{1}{\sqrt{2πk}} (1 + \frac{1}{12k} + O(\frac{1}{k^{2}} )).

In particular, for k = 1, the ratio of the approximation of (2.3) to the usual saddlepoint approximation is close to \sqrt{2π}/e = 0.922. So even for k = 1 the usual saddlepoint approximation based on a normal approximation for the tilted mean is quite good.

3 Hypergeometric Distribution

Put S_{1} = ∑_{i=1}^{N} a_{i}I_{i} and S_{2} = ∑_{i=1}^{N} I_{i} where

\{ \begin{matrix}a_{i} = 1 & i = 1, …, m, \\ a_{i} = 0 & i = m + 1, …, N \end{matrix}

and

\{ \begin{matrix}P (I_{i} = 1) = \frac{n}{N} = p, \\ P (I_{i} = 0) = 1 − p = q & i = 1, …, N. \end{matrix}

X_{1}, …, X_{n} are a sample from a_{1}, …, a_{N}. Then the hypergeometric variable H = X_{1} + … + X_{n} and

P (H = k) = P (S_{1} = k|S_{2} = n).

We have S_{1} = B_{1}, S_{2} = B_{1} + B_{2} where B_{1} = ∑_{i=1}^{m} I_{i}, B_{2} = ∑_{i=m+1}^{N} I_{i}. Then, as binomial situation, define tilted random variable X_{1t},…,X_{Nt} as

f_{t}(x) = P (X_{1t} = x) = \frac{e^{tx}p^{x}q^{1−x}}{q + pe^{tx}} = e^{−K(t)+tx}f(x),  (3.5)

where f(x) = p^{x}q^{1−x}, x=0,1 and

K(t) = q + pe^{tx}.

Let B_{1τ} = ∑_{i=1}^{m} I_{iτ}, B_{2η} = ∑_{i=m+1}^{N} I_{iη} and S_{2t} = ∑_{i=1}^{N} I_{it}. Then

P (H = k) = \frac{P (B_{1} = k, B_{1} + B_{2} = n)}{P (S_{2} = n)} = \frac{P (B_{1} = k)P (B_{2} = n − k)}{P (S_{2} = n)}
= e^{−mΛ(k/m)−(N−m)Λ((n−k)/(N−m))+NΛ(n/N)} \frac{P (B_{1τ} = k)P (B_{2η} = n − k)}{P (S_{2t} = n)}.

From K′(τ) = k/m, K′(η) = (n − k)/(N − m) and K′(t) = n/N we can get τ = τ(x), η = η(x) and t = t(x) respectively where x = k/n. As in the binomial situation, we know B_{1τ} is binomial (m, k/m), B_{2η} is binomial (N − m, (n − k)/(N − m)) and S_{2t} is binomial (n, n/N). We consider n < min(m, N −m) situation only in the following theorem. Other situations can be handled similarly.

Theorem 3.1 For H a hypergeometric variable, for n < min(m, N − m) and N → ∞ we have

(a) for k → ∞, k ≤ n/2,

P (H = nx) = e^{−NΛ_{1}(x)}\sqrt{\frac{q}{2πnx(1 − x)(1 − px/p′)(1 − p(1 − x)/q′)}}(1 + O(\frac{1}{k′})), (3.6)

where p′ = m/N = 1 − q′, k′ = min(k, N − m − n) and

Λ_{1}(x) = p′Λ(px/p′) + q′Λ(p(1 − x)/q′) − NΛ(p).

(b) for k = 0,

P (H = 0) = e^{−NΛ_{1}(0)}\sqrt{\frac{qq′}{q′ − p}}(1 + O(\frac{1}{n′})) = \frac{q^{N−n+1/2}q^{′N−m+1/2}}{(q′ − p)^{N−m−n+1/2}} (1 + O(\frac{1}{n′})), (3.7)

where n′ = min(n, N − m − n).

(c) for k a fixed integer,

P (H = nx) = e^{−NΛ_{1}(x)} \frac{k^{k}e^{−k}}{k!} \sqrt{\frac{qq′}{q′ − p}}(1 + O(\frac{1}{n′})).  (3.8)

Proof: (a) The proof of the result follows from Theorem 2.1.

(b) When k = 0, we have P (B_{1τ} = 0) = 1 and

P (H = 0) = exp\{−N lim_{x→0} Λ_{1}(x)\}\frac{P (B_{1τ} = 0)P (B_{2η} = n)}{P (S_{2} = n)}
= exp\{−N lim_{x→0} Λ_{1}(x)\}\sqrt{\frac{n(1 − p)}{n(1 − \frac{n}{(N−m)})}}(1 + O(\frac{1}{k′}))
= \frac{(1 − p)^{N−n}(1 − p′)^{N−m}}{(1 − p′ − p)^{N−m−n}} \sqrt{\frac{(1 − p′)(1 − p)}{1 − p′ − p}}(1 + O(\frac{1}{k′}))
= \frac{q^{N−n+1/2}q^{′N−m+1/2}}{(q′ − p)^{N−m−n+1/2}} (1 + O(\frac{1}{k′})),

where k′ = min(n, N − n − m).

(c) For k a fixed integer, the tilted variable B_{1τ} is binomial (m, k/m). Since m∗k/m = k is fixed, B_{1τ} converges in distribution to a Poisson(k) variable. The proof of the result follows from Theorem 2.1. □

4 One-sample Wilcoxon Statistic

Put W = ∑_{j=1}^{n} jI_{j} where P (I_{j} = 0) = P (I_{j} = 1) = \frac{1}{2}, j = 1, …, n. So, the moment generating function of W is

M(t) = ∏_{j=1}^{n}(\frac{1}{2} + \frac{1}{2}e^{jt}).

Put K(t) = \frac{1}{n} ∑_{j=1}^{n} log(\frac{1}{2} + \frac{1}{2}e^{jt}). Then the density functions of the tilted variables I_{jt}, j = 1, …, n are

P (I_{jt} = 1) = \frac{e^{jt}}{1 + e^{jt}} , P (I_{jt} = 0) = \frac{1}{1 + e^{jt}} , j = 1, …, n.  (4.9)

Put W_{t} = ∑_{j=1}^{n} jI_{jt}. Then,

f_{W_{t}}(nw) = e^{ntw−nK_{n}(t)}f_{W} (nw).

From

K′(t) = \frac{1}{n} ∑_{j=1}^{n} \frac{je^{jt}}{1 + e^{jt}} = w = k/n,  (4.10)

we can get saddlepoint t = t_{n}(k). Put Λ(w) = wt(w) − K(t(w)). Then

f_{W} (nw) = e^{−nΛ(w)}f_{W_{t}}(nw).  (4.11)

Put h_{n}(t) = nK′(t) = ∑_{j=1}^{n} \frac{je^{jt}}{1+e^{jt}}. Then h′_{n}(t) = nK′′(t) = ∑_{j=1}^{n} \frac{j^{2}e^{jt}}{(1+e^{jt})^{2}} > 0. So, h_{n}(t) is an increasing function. We can get the following Lemma.

Lemma 4.1 The solution t_{n} of the equation

h_{n}(t) = ∑_{j=1}^{n} \frac{je^{jt}}{1 + e^{jt}} = k

satisfies

\{ \begin{matrix}t_{n}(k) = −∞ & for k = 0, \\ t_{n}(k) < 0 & for 0 < k < n(n + 1)/4, \\ t_{n}(k) = 0 & for k = n(n + 1)/4, \\ t_{n}(k) > 0 & for n(n + 1)/4 < k < n(n + 1)/2, \\ t_{n}(k) = ∞ & for k = n(n + 1)/2. \end{matrix}  (4.12)

and t_{n}(k) is an increasing function with respect to k as n is fixed.

Note t = t(w) = t_{n}(k). Without losing generality, we consider k ≤ n(n + 1)/4 only from now on.

Lemma 4.2 For 1 ≤ k ≤ n(n + 1)/c, where c > 4, the solution t_{n}(k) of the equation

h_{n}(t) = ∑_{j=1}^{n} \frac{je^{jt}}{1 + e^{jt}} = k  (4.13)

satisfies

−t_{n}(k) ∼ \frac{1}{\sqrt{k}} , as n → ∞,

where this means

\frac{C_{1}}{\sqrt{k}} < −t_{n}(k) < \sqrt{\frac{1 + \sqrt{5}}{2}} \frac{1}{\sqrt{k}} ,

where C_{1} = e^{−2/\sqrt{k}}((−c_{1} − 1)e^{−c_{1}} + 1)/2 > 0, where c_{1} = log(c/4) > 0. Also, for k < n(n + 1)/4,

−t_{n}(k) = O(\frac{1}{\sqrt{k}}), as n → ∞.

Proof: Put t = t_{n}(k). We know t < 0. Therefore,

∫_{0}^{1} ye^{nt(y+ \frac{1}{n})}dy < \frac{1}{n^{2}} ∑_{j=1}^{n} je^{jt} = ∑_{j=1}^{n} \frac{j}{n}e^{\frac{j}{n} nt} \frac{1}{n} < ∫_{0}^{1}(y + \frac{1}{n})e^{nty}dy.

So,

e^{t}n^{2} ∫_{0}^{1} ye^{nty}dy < ∑_{j=1}^{n} je^{jt} < n^{2} ∫_{0}^{1} ye^{nty}dy + n ∫_{0}^{1} e^{nty}dy.

Now

n^{2} ∫_{0}^{1} ye^{nty}dy = ∫_{0}^{−nt} ze^{−z}dz \frac{1}{t^{2}} = ((nt − 1)e^{nt} + 1)\frac{1}{t^{2}} ,

n ∫_{0}^{1} e^{nty}dy = ∫_{0}^{−nt} e^{−z}d(−z)\frac{1}{t} = (e^{nt} − 1)\frac{1}{t},

and 1 < 1 + e^{jt} < 2. So,

e^{t}((nt − 1)e^{nt} + 1)\frac{1}{2t^{2}} < k = ∑_{j=1}^{n} \frac{je^{jt}}{1 + e^{jt}}
< ((nt − 1)e^{nt} + 1 + t(e^{nt} − 1))\frac{1}{t^{2}} .

We know

\frac{n(n + 1)}{c} > k = ∑_{j=1}^{n} \frac{je^{jt}}{1 + e^{jt}} > ∑_{j=1}^{n} \frac{je^{nt}}{1 + e^{jt}} > \frac{1}{2}e^{nt} ∑_{j=1}^{n} j = \frac{n(n + 1)}{4}e^{nt}.

That is e^{nt} < \frac{4}{c} = e^{−c_{1}} < 1 where c_{1} > 0. So,

nt < −c_{1} < 0.

Also f(x) = (x − 1)e^{x} + 1 is a decreasing function when x ≤ 0. So,

0 = f(0) < (−c_{1} − 1)e^{−c_{1}} + 1 < (x − 1)e^{x} + 1 ≤ 1

when x < −c_{1}. Therefore we have

e^{t}((−c_{1} − 1)e^{−c_{1}} + 1)\frac{1}{2k} < t^{2} < (1 − t)\frac{1}{k}.  (4.14)

When k = 1 we have t^{2} + t − 1 < 0. So, −(1 + \sqrt{5})/2 < t_{n}(1) < 0. From Lemma 4.1, we know t_{n}(k) is increasing function with respect to k. So,

t_{n}(1) < t_{n}(k) for k > 1.

Therefore, from (4.14),

\frac{C_{1}}{\sqrt{k}} < −t_{n}(k) < \sqrt{\frac{1 + \sqrt{5}}{2}} \frac{1}{\sqrt{k}} as n → ∞,

where C_{1} = \sqrt{e^{−2/\sqrt{k}}((−c_{1} − 1)e^{−c_{1}} + 1)/2}. Also because t_{n}(k) is an increasing function with respect to k, we have

−t_{n}(k) = O(\frac{1}{\sqrt{k}}), as n → ∞

when 1 ≤ k ≤ n(n + 1)/4. □

Lemma 4.3 If k is fixed, we have

t_{n}(k) → t(k) as N → ∞,

where t(k) < 0 is dependent on k only.

Proof: We have

h_{n}(t_{n}(k)) = ∑_{j=1}^{n} \frac{je^{jt_{n}(k)}}{1 + e^{jt_{n}(k)}} = k

and

∑_{j=1}^{n} \frac{je^{jt_{n+1}(k)}}{1 + e^{jt_{n+1}(k)}} = k − \frac{(n + 1)e^{(n+1)t_{n+1}(k)}}{1 + e^{(n+1)t_{n+1}(k)}} < k.

We know h_{n}(t) is an increasing function. Therefore,

t_{n+1}(k) ≤ t_{n}(k) < 0.

From Lemma 4.2, we know t_{n}(k) is absolutely bounded for any n. Therefore,

t_{n}(k) → t(k) as N → ∞. □

Consider the following 3 cases.

(a) k → ∞ and n(n + 1)/2 − k → ∞

(b) k = 0(or n),

(c) k is fixed and finite integer (or n(n + 1)/2 − k is fixed and finite integer). In each of these cases we have a different limiting distribution for W_{t(w)}.

Theorem 4.1 Assume W = ∑_{j=1}^{n} jI_{j} where P (I_{j} = 0) = P (I_{j} = 1) = \frac{1}{2}, j = 1, …, n. Then, as n → ∞ we have

(a) for k → ∞, k ≤ n(n + 1)/4,

P (W = k) = e^{−nΛ(k/n)} \frac{1}{\sqrt{2πK′′(t_{n}(k))}}(1 + O(k^{−1/2}));  (4.15)

(b) for k = 0,

P (W = 0) = e^{−nΛ(0)} = 2^{−n};

(c) for k a fixed number and k ≤ n,

P (W = k) = e^{−nΛ(k/n)}f_{W_{t(k)}}(k)(1 + O(ne^{nt(k)})),  (4.16)

where t(k) = lim_{n→∞} t_{n}(k), f_{W_{t(k)}}(k) = B(k)e^{kt(k)}/(∏_{j=1}^{∞}(1 + e^{jt(k)})) and B(k) is the number of sets of distinct positive integers whose sum is k.

Proof: (a) When k → ∞ and k ≤ n(n + 1)/4, note −t = O(k^{−1/2}) from Lemma 4.2, we have

\{ \begin{matrix}K′′(t) = n^{2} \frac{1}{n} ∑_{j=1}^{n} \frac{( \frac{j}{n})^{2}e^{\frac{j}{n} nt}}{(1+e^{\frac{j}{n} nt})^{2}} ∼ n^{2} ∑_{j=1}^{n}(\frac{j}{n})^{2}e^{\frac{j}{n} nt} \frac{1}{n} ∼ n^{2} ∫_{0}^{1} y^{2}e^{ynt}dy ∼ n^{−1}k^{3/2}, \\ K^{(3)}(t) = n^{3} \frac{1}{n} ∑_{j=1}^{n} \frac{( \frac{j}{n})^{3}e^{\frac{j}{n} nt}(1−e^{\frac{j}{n} nt})}{(1+e^{\frac{j}{n} nt})^{3}} ∼ n^{3} ∑_{j=1}^{n}(\frac{j}{n})^{3}e^{\frac{j}{n} nt} \frac{1}{n} ∼ n^{3} ∫_{0}^{1} y^{2}e^{ynt}dy ∼ n^{−1}k^{2}, \\ K^{(4)}(t) ∼ n^{−1}k^{5/2}, \\ K^{(5)}(t) ∼ n^{−1}k^{3} \end{matrix}

(4.17)

So nK′′(t) ∼ k^{3/2} → ∞. We know

P (W_{t} = k) = \frac{1}{2π} ∫_{−π}^{π} exp\{−iξk + nK(t + iξ) − nK(t)\}dξ = J_{1} + J_{2},

where

J_{1} = \frac{1}{2π} ∫_{−ϵk^{−1/2}}^{ϵk^{−1/2}} exp\{−iξk + nK(t + iξ) − nK(t)\}dξ

and

J_{2} = \frac{1}{2π} ∫_{(−π,π)\(−ϵk^{−1/2},ϵk^{−1/2})} exp\{−iξk + nK(t + iξ) − nK(t)\}dξ.

Now we look at J_{1} first.

J_{1} = \frac{1}{2π} ∫_{−ϵk^{−1/2}}^{ϵk^{−1/2}} exp\{−\frac{1}{2}nK′′(t)ξ^{2}+\frac{nK^{(3)}(t)}{3!}(iξ)^{3}+\frac{nK^{(4)}(t)}{4!}(iξ)^{4}+\frac{nK^{(5)}(t + θiξ)}{5!}(iξ)^{5}\}dξ.

Put η = \sqrt{nK′′(t)}ξ. Then, from

\frac{nK^{(3)}(t)}{(nK′′(t))^{3/2}} ∼ \frac{1}{k^{1/4}} , \frac{nK^{(4)}(t)}{(nK′′(t))^{2}} ∼ \frac{1}{k^{1/2}} , \frac{K^{(5)}(t)}{(K′′(t))^{5/2}} ∼ \frac{1}{k^{3/4}} ,

when ϵ > 0 is small enough, we have

J_{1} = \frac{1}{2π\sqrt{nK′′(t)}} ∫_{−ϵk^{−1/2}\sqrt{nK′′(t)}}^{ϵk^{−1/2}\sqrt{nK′′(t)}} e^{− \frac{1}{2} η^{2}}(1 + \frac{K^{(3)}(t)}{3!K′′(t)^{3/2}\sqrt{n}}(iη)^{3} + \frac{K^{(4)}(t)}{4!K′′(t)^{2}n}(iη)^{4}

+\frac{K^{(3)}(t)^{2}}{2!(3!)^{2}K′′(t)^{3}n}(iη)^{6} + P_{5}(η)e^{\frac{1}{4} η^{2}})dη,

where P_{5}(η) is the polynomial of degree 5. Because

\frac{K^{(3)}(t)}{(K′′(t))^{3/2}\sqrt{n}} ∼ \frac{1}{k^{1/4}} , \frac{K^{(4)}(t)}{(K′′(t))^{2}n} ∼ \frac{1}{k^{1/2}} ,

we have

J_{1} = \frac{1}{\sqrt{2πnK′′(t)}}(1 + O(\frac{1}{k^{1/2}} )).

Now let us look at J_{2}. We use a proof related to one used in Albers, W., Bickel, P. J. and van Zwet, W. R. (1976). J_{2} is rewritten as

J_{2} = \frac{1}{2π} ∫_{(−π,π)\(−ϵk^{−1/2},ϵk^{−1/2})} e^{−iξl} ∏_{j=1}^{n} \frac{1 + e^{jt+ijξ}}{1 + e^{jt}} dξ.  (4.18)

When ϵk^{−1/2} ≤ |ξ| ≤ π where ϵ is small enough, note t ∼ −k^{−1/2}, we have

|∏_{j=1}^{n} \frac{1 + e^{jt+ijξ}}{1 + e^{jt}} | = ∏_{j=1}^{n} \frac{((1 + e^{jt})^{2} − 2e^{jt}(1 − cos(jξ)))^{1/2}}{(1 + e^{t})}

= ∏_{j=1}^{n} (1 − 4\frac{e^{jt}}{(1 + e^{jt})^{2}} sin^{2}(jξ/2))^{1/2} ≤ exp(−c_{1} ∑_{j=1}^{[k^{1/2}]} sin^{2}(jξ/2)),

where c_{1} > 0. For ϵk^{−1/2} ≤ |ξ| ≤ ϵ, then |ξ| = ϵk_{1}^{−1/2} where 1 ≤ k_{1} ≤ k. Separating the integers 1, …, [k^{1/2}] to 1, …, [k_{1}^{1/2}]; [k_{1}^{1/2}]+1, …, 2[k_{1}^{1/2}] and so on, we can get [[k^{1/2}]/[k_{1}^{1/2}]] groups. In at least half of the groups at least one third of the j’s satisfy

|jξ/2 − 2Kπ| > ϵ/4 for all K.

So,

exp(−c_{1} ∑_{j=1}^{[k^{1/2}]} sin^{2}(jξ/2)) ≤ exp(−\frac{c_{1}k^{1/2}}{6} sin^{2}(ϵ/4)).

For ϵ ≤ |ξ| ≤ π, we can find k_{2} ≥ [k^{1/2}/2] values of j such that for each of these

|jξ/2 − 2Kπ| > δ > 0 for all K

since in this case two consecutive j’s cannot both be within δ of multiples of 2π. So,

∑_{j=1}^{[k^{1/2}]} c_{1}sin^{2}(jξ/2) ≥ c_{0}k^{1/2} for all ϵk^{−1/2} ≤ |ξ| ≤ π

for some c_{0} > 0. Therefore,

|∏_{j=1}^{n} \frac{1 + e^{jt+ijξ}}{1 + e^{jt}} | ≤ e^{−c_{0}k^{1/2}}.

So, from (4.18), we know

J_{2} = O(e^{−c_{0}k^{1/2}}) = O(\frac{1}{k^{3/2}} ).

Therefore,

P (W = k) = e^{−nΛ(k/n)} \frac{1}{\sqrt{2πK′′(t_{n}(k))}}(1 + O(k^{−1/2}))

for k → ∞ and k ≤ n(n + 1)/4.

(b) for k = 0, we know t = −∞. So,

lim_{x→0}(2^{n} exp(−nΛ(x))) = lim_{t→−∞} ∏_{j=1}^{n}(1 + e^{jt}) = 1

and ,from (4.9),

f_{W_{t}}(0) = lim_{t→−∞} ∏_{j=1}^{n} \frac{1}{1 + e^{jt}} = 1.

From (4.11), we have

f_{W} (0) = exp(−n lim_{x→0} Λ(x))f_{W_{t}}(0) = 2^{−n}.

(c) When k is fixed and k ≤ n, we have

e^{−nΛ(k/n)} = 2^{−n}e^{−kt_{n}(k)} ∏_{j=1}^{n}(1 + e^{jt_{n}(k)}).  (4.19)

From Lemma 4.2, we have t_{n}(k) → t(k) < 0. We also can prove t_{n}(k) = t(k)(1+O(ne^{nt(k)})) for some c > 0. Let B_{n}(k) and B(k) be the number of sets of distinct positive integers from {1, …, n} and {1, 2, …} respectively whose sums are k. From W_{t} = ∑_{j=1}^{n} jI_{jt}, where P (I_{jt} = 1) = e^{jt}/(1 + e^{jt}), when k ≤ n, we have

f_{W_{tn(k)}}(k) = \frac{B_{n}(k)e^{kt_{n}(k)}}{∏_{j=1}^{n}(1 + e^{jt_{n}(k)})} = \frac{B(k)e^{kt(k)}}{∏_{j=1}^{∞}(1 + e^{jt(k)})}(1+O(ne^{nt(k)})) = f_{W_{t(k)}}(k)(1+O(ne^{nt(k)})),

(4.20)

where t(k) and f_{W_{t(k)}}(k) depend only on k, and (4.16) follows. □

Remark: Table 1 gives values of B(k), t(k), exact values of P (W = k) and approximations of P (W = k) based on (4.16) and (4.15) for k = 10. From these we see that the error O(ne^{nt(k)}) is not small unless n is large and k is small. So, for k = 8, 9, 10, the new approximations begin to fail, however , it is greatly superior to the normal approximations for k ≤ 5.

Table 1 Approximations for one sample Wilcoxon for n = 10

k  1  2  3  4  5  6  7  8  9  10
B_{10}(k) = B(k) 1  1  2  2  3  4  5  6  8  10
t(k)  -.889 -.635 -.520 -.451 -.404 -.369 -.342 -.320 -.302 -.286
exact  .0010 .0010 .0020 .0020 .0029 .0039 .0049 .0059 .0078 .0098
P_{new}(W = k) .0010 .0010 .0019 .0019 .0028 .0037 .0044 .0051 .0065 .0077
P_{nor}(W = k) .0012 .0015 .0020 .0026 .0034 .0043 .0054 .0067 .0082 .0098

4.1 Two-sample Wilcoxon statistic

Put S_{1} = ∑_{i=1}^{N} jI_{j}, S_{2} = ∑_{j=1}^{N} I_{j} where

P (I_{j} = 1) = \frac{m}{n + m} = \frac{m}{N} = p, P (I_{j} = 0) = \frac{n}{N} = 1 − p = q, i = 1, …, N.

Put U_{m,n} is the two-sample Wilcoxon statistic. Then

P (U_{m,n} = k) = P (S_{1} = k|S_{2} = m) = \frac{P (S_{1} = k, S_{2} = m)}{P (S_{2} = m)}.  (4.21)

The moment generating function of S_{1} and S_{2} is

M(s, t) = Ee^{sS_{1}+tS_{2}} = ∑_{j=1}^{N} Ee^{(js+t)I_{j}} = ∏_{j=1}^{N}(q + pe^{js+t}).

Put K(s, t) = \frac{1}{N} ∑_{j=1}^{N} log(q + pe^{js+t}). Then the density functions of the tilted variables I_{jst}, j = 1, …, N are

P (I_{jst} = 1) = \frac{pe^{js+t}}{q + pe^{js+t}} , P (I_{jst} = 0) = \frac{q}{q + pe^{js+t}} , j = 1, …, N.  (4.22)

Put S_{1st} = ∑_{j=1}^{N} jI_{jst} and S_{2st} = ∑_{j=1}^{N} I_{jst} Then,

P (S_{1} = k, S_{2} = m) = e^{NK(s,t)−sk−tm}P (S_{1st} = k, S_{2st} = m).

From

\{ \begin{matrix}NK_{10}(s, t) = ∑_{j=1}^{N} \frac{jpe^{js+t}}{q+pe^{js+t}} = k \\ NK_{01}(s, t) = ∑_{j=1}^{N} \frac{pe^{js+t}}{q+pe^{js+t}} = m, \end{matrix}  (4.23)

where K_{ij}(s, t) = ∂^{i+j}K(s, t)/(∂s^{i}∂t^{j}), we can get saddlepoint s = s(k, m), t = t(k, m). Put Λ(k/N, m/N) = \frac{1}{N} (s(k, m)k + t(k, m)m − K(s(k, m), t(k, m))). Then

P (S_{1} = k, S_{2} = m) = e^{−NΛ(k/N,m/N)}P (S_{1st} = k, S_{2st} = m).  (4.24)

Partially differentiating (4.23) with respect to k, we can get

\{ \begin{matrix}K_{20}(s, t)s_{10}(k, m) + K_{11}(s, t)t_{10}(k, m) = \frac{1}{N} \\ K_{11}(s, t)s_{10}(k, m) + K_{02}(s, t)t_{10}(k, m) = 0. \end{matrix}

So,

\{ \begin{matrix}s_{10}(k, m) = \frac{1}{N} \frac{K_{20}(s,t)}{K_{02}(s,t)K_{20}(s,t)−K_{11}^{2}(s,t)} > 0 \\ t_{10}(k, m) = \frac{1}{N} \frac{−K_{11}(s,t)}{K_{02}(s,t)K_{20}(s,t)−K_{11}^{2}(s,t)} < 0. \end{matrix}  (4.25)

Therefore we can get the following Lemma.

Lemma 4.4 Put k = m(m + 1)/2 + l where 0 ≤ l ≤ \frac{m(2N−m+1)}{2} − \frac{m(m+1)}{2} = mn. The solution s and t of equation (4.23) satisfies

s → −∞, ms + t → ∞, (m + 1)s + t → −∞,  as l → 0,
s < 0, t > 0  as 0 < l < \frac{1}{2}mn,
s = t = 0  as l = \frac{1}{2}mn, when mn is even,
s > 0, t < 0  as \frac{1}{2}mn < l < mn,
s → ∞, (N − m)s + t → −∞, (N − m + 1)s + t → ∞, as l → mn

(4.26)

If N, p are fixed, s(k, m) is an increasing function with respect to k and t(k, m) is an decreasing function with respect to k.

We consider 0 < l < \frac{1}{2}mn first. In this situation s < 0 and t > 0 from Lemma 4.4. So, from (4.23),

N ∫_{0}^{1} \frac{e^{(y+1/N)Ns+t}}{q + pe^{(y+1/N)Ns+t}} dy < ∑_{j=1}^{N} \frac{e^{\frac{j}{N} s+t}}{q + pe^{\frac{j}{N} s+t}} \frac{1}{N} N = N < N ∫_{0}^{1} \frac{e^{yNs+t}}{q + pe^{yNs+t}} dy.

That is

\frac{1}{ps}ln\frac{q + pe^{Ns+s+t}}{q + pe^{s+t}} < N < \frac{1}{ps}ln\frac{q + pe^{Ns+t}}{q + pe^{t}} .

By doing some derivation we can get

\frac{q}{p} \frac{1 − e^{ms}}{1 − e^{ns}} e^{−ms} < e^{t} < \frac{q}{p} \frac{1 − e^{ms}}{1 − e^{(n−1)s}} e^{−(m+1)s}.  (4.27)

From (4.23), we have

\{ \begin{matrix}f_{1N}(s, t) = ∑_{j=1}^{N} \frac{jpe^{js+t}}{q+pe^{js+t}} − \frac{m(m+1)}{2} = − ∑_{j=1}^{m} \frac{jq}{q+pe^{js+t}} + ∑_{j=m+1}^{N} \frac{jpe^{js+t}}{q+pe^{js+t}} = l \\ f_{2N}(s, t) = ∑_{j=1}^{N} \frac{pe^{js+t}}{q+pe^{js+t}} − m = − ∑_{j=1}^{m} \frac{q}{q+pe^{js+t}} + ∑_{j=m+1}^{N} \frac{pe^{js+t}}{q+pe^{js+t}} = 0 \end{matrix} (4.28)

If Ns → 0 we can get −ms ≤ t ≤ −(m + 1)s from (4.27). So,

∑_{j=1}^{N} \frac{jpe^{js+t}}{q + pe^{js+t}} − \frac{m(m + 1)}{2} = \frac{mn}{2}(1 + ϵ(N))

where ϵ(N) → 0 as N → ∞. Therefore we have Lemma 4.5 If 0 < l < mn/c where c > 2, then −Ns > c_{1} > 0. Without loss of generality, we assume m ≥ n. Lemma 4.6 If 0 < l < mn/2 and p → p_{0}, then

t = −ms − δs + log(q/p) + log c_{3}

where 0 ≤ δ ≤ 1 and 1 ≤ c_{3} ≤ m/(n − 1). Proof: Put a = m/(n − 1) > 1 and x = e^{(n−1)s}. We know s < 0 for 0 < l < mn/2. So, 0 < x < 1. Put

f(x) = \frac{1 − x^{a}}{1 − x} = \frac{1 − e^{ms}}{1 − e^{(n−1)s}} .

Then,

f′(x) = \frac{(a − 1)x^{a} − ax^{a−1} + 1}{(1 − x)^{2}} = \frac{1}{(1 − x)^{2}} g(x),  (4.29)

where g(x) = (a − 1)x^{a} − ax^{a−1} + 1. We have

g′(x) = a(a − 1)x^{a−2}(1 − x) > 0

and lim_{x→0} g(x) = 1. So, f′(x) > 0. From lim_{x→0} f(x) = 1 and lim_{x→1} f(x) = a, we have 1 ≤ f(x) ≤ a. In the same way, we can prove 1 ≤ (1 − e^{ms})/(1 − e^{ns}) ≤ m/n. From (4.27), the proof of the Lemma is now complete. □ Lemma 4.7 If 0 < l < mn/c where c > 2 and p → p_{0}, then, there exist constants c_{5} > 0 and c_{6} < 3 + \sqrt{13} such that

\frac{c_{5}}{\sqrt{l}} < −s < \frac{c_{6}}{\sqrt{l}}, as N → ∞

for large enough N. Also, for mn/c ≤ l < mn/2,

−s = O(\frac{1}{\sqrt{l}}), as N → ∞.

Proof: From (4.28), we have

\{ \begin{matrix}f_{1N}(s, t) = − ∑_{j=1}^{m} \frac{j}{1+c_{3}e^{js−ms−δs}} + ∑_{j=m+1}^{N} \frac{jc_{3}e^{js−ms−δs}}{1+c_{3}e^{js−ms−δs}} = l, \\ f_{2N}(s, t) = − ∑_{j=1}^{m} \frac{1}{1+c_{3}e^{js−ms−δs}} + ∑_{j=m+1}^{N} \frac{c_{3}e^{js−ms−δs}}{1+c_{3}e^{js−ms−δs}} = 0. \end{matrix}  (4.30)

That is

\{ \begin{matrix}f_{1N}(s, t) = − ∑_{j=1}^{m} \frac{je^{(m−j)s+δs}}{c_{3}+e^{(m−j)s+δs}} + ∑_{j=1}^{n} \frac{(m+j)c_{3}e^{js−δs}}{1+c_{3}e^{js−δs}} = l, \\ f_{2N}(s, t) = − ∑_{j=1}^{m} \frac{e^{(m−j)s+δs}}{c_{3}+e^{(m−j)s+δs}} + ∑_{j=1}^{n} \frac{c_{3}e^{js−δs}}{1+c_{3}e^{js−δs}} = 0. \end{matrix}  (4.31)

Therefore

f_{N}(s, t) = f_{1N}(s, t)−mf_{2N}(s, t) = ∑_{j=1}^{m−1} \frac{je^{js+δs}}{c_{3} + e^{js+δs}} +∑_{j=1}^{n−1} \frac{(j + 1)c_{3}e^{js+(1−δ)s}}{1 + c_{3}e^{js+(1−δ)s}} +\frac{c_{3}e^{(1−δ)s}}{1 + c_{3}e^{(1−δ)s}} = l.

(4.32)

We know s < 0 and f_{1}(x) = x/(1 + x) is an increasing function. So

∫_{0}^{1−1/m} ye^{ms(y+ \frac{1+δ}{m})}dy < \frac{1}{m^{2}} ∑_{j=1}^{m−1} je^{js+δs} = ∑_{j=1}^{m−1} \frac{j}{m}e^{\frac{j+δ}{m} ms} \frac{1}{m} < ∫_{δ/m}^{(m−1+δ)/m}(y+\frac{1 − δ}{m})e^{msy}dy.

That is

m^{2}e^{(1+δ)s} ∫_{0}^{1−1/m} ye^{msy}dy < ∑_{j=1}^{m−1} je^{js+δs} < m^{2} ∫_{δ/m}^{(m−1+δ)/m} ye^{msy}dy+m ∫_{δ/m}^{(m−1+δ)/m}(1−δ)e^{msy}dy

Now

\{ \begin{matrix}m^{2} ∫_{0}^{1−1/m} ye^{msy}dy = \{((m − 1)s − 1)e^{(m−1)s} + 1\}\frac{1}{s^{2}} , \\ m^{2} ∫_{δ/m}^{(m−1+δ)/m} ye^{msy}dy = \{((m − 1 + δ)s − 1)e^{(m−1+δ)s} + (1 − δs)e^{δs}\}\frac{1}{s^{2}} , \\ m ∫_{δ/m}^{(m−1+δ)/m}(1 − δ)e^{msy}dy = (1 − δ)(e^{(m−1+δ)s} − e^{δs})\frac{1}{s} \end{matrix}  (4.33)

and 1 < c_{3} + e^{js+δs} < 1 + c_{3} as j = 1, …, m − 1. So,

e^{(1+δ)s}\{((m − 1)s − 1)e^{(m−1)s} + 1\}\frac{1}{(1 + c_{3})s^{2}} < ∑_{j=1}^{m−1} \frac{je^{js+δs}}{c_{3} + e^{js+δs}} < (mse^{(m−1)s} + (1 − s))\frac{e^{δs}}{s^{2}} .

We know −1 ≤ mse^{(m−1)s} < 0 and −1 ≤ (m − 1)s − 1)e^{(m−1)s} < 0. Therefore,

e^{(1+δ)s}\{((m − 1)s − 1)e^{(m−1)s} + 1\}\frac{1}{(1 + c_{3})s^{2}} < ∑_{j=1}^{m−1} \frac{je^{js+δs}}{c_{3} + e^{js+δs}} < (1 − s)\frac{e^{δs}}{s^{2}} . (4.34)

In the same way we can get

∫_{0}^{1−1/n}(y + 1/n)e^{ns(y+ \frac{2−δ}{n})}dy < \frac{1}{n^{2}} ∑_{j=1}^{n−1}(j + 1)e^{js+(1−δ)s} < ∫_{(1−δ)/n}^{(n−δ)/n}(y + \frac{1 + δ}{n})e^{nsy}dy.

That is

n^{2}e^{(2−δ)s} ∫_{0}^{1−1/n} ye^{nsy}dy + ne^{(2−δ)s} ∫_{0}^{1−1/n} e^{msy}dy < ∑_{j=1}^{n−1} je^{js+δs}
< n^{2} ∫_{(1−δ)/n}^{(n−δ)/n} ye^{nsy}dy + n ∫_{(1−δ)/n}^{(n−δ)/n}(1 + δ)e^{nsy}dy.

Now

\{ \begin{matrix}n^{2} ∫_{0}^{1−1/n} ye^{nsy}dy = \{((n − 1)s − 1)e^{(n−1)s} + 1\}\frac{1}{s^{2}} , \\ n ∫_{0}^{1−1/n} e^{nsy}dy = (e^{(n−1)s} − 1)\frac{1}{s}, \\ n^{2} ∫_{(1−δ)/n}^{(n−δ)/n} ye^{nsy}dy = \{((n − δ)s − 1)e^{(n−δ)s} + (1 − (1 − δ)s)e^{(1−δ)s}\}\frac{1}{s^{2}} , \\ n ∫_{(1−δ)/n}^{(n−δ)/n}(1 + δ)e^{nsy}dy = (1 + δ)(e^{(n−δ)s} − e^{(1−δ)s})\frac{1}{s} \end{matrix}  (4.35)

and 1 < 1 + c_{3}e^{js−δs} < 1 + c_{3} as j = 1, …, n. So,

e^{(2−δ)s}((ns − 1)e^{(n−1)s} − s + 1)\frac{c_{3}}{(1 + c_{3})s^{2}} < ∑_{j=1}^{n−1} \frac{(j + 1)c_{3}e^{js+(1−δ)s}}{1 + c_{3}e^{js+(1−δ)s}}
< c_{3}((n + 1)se^{(n−1)s} + (1 − 2s))\frac{e^{(1−δ)s}}{s^{2}} .

We know −1 ≤ (n + 1)se^{(n−1)s} < 0 and −1 ≤ (ns − 1)e^{(n−1)s} < 0. Therefore,

e^{(2−δ)s}((ns − 1)e^{(n−1)s} − s + 1)\frac{c_{3}}{(1 + c_{3})s^{2}} < ∑_{j=1}^{n−1} \frac{(j + 1)c_{3}e^{js+(1−δ)s}}{1 + c_{3}e^{js+(1−δ)s}} < c_{3}(1 − 2s)\frac{e^{(1−δ)s}}{s^{2}} .

(4.36)

Put δ_{1} = c_{3}e^{(1−δ)s}/(1 + c_{3}e^{(1−δ)s}). We know 0 < δ_{1} ≤ c_{3}/(1 + c_{3}). So, substituting (4.34), (4.36) and δ_{1} in (4.32), we get

e^{2s}\{(ns − 1)e^{(n−1)s} + ((m − 1)s − 1)e^{(m−1)s} − s + 2\}\frac{1}{(1 + c_{3})s^{2}} < l − δ_{1} < c_{3}(2 − 3s)\frac{1}{s^{2}} .

That is

e^{2s}\{(ns − 1)e^{(n−1)s} + ((m − 1)s − 1)e^{(m−1)s} − s + 2\}\frac{1}{(1 + c_{3})(l − δ_{1})} < s^{2} < \frac{c_{3}(2 − 3s)}{l − δ_{1}}.

From Lemma 4.5, we can prove

(ns − 1)e^{(n−1)s} + ((m − 1)s − 1)e^{(m−1)s} − s + 2 > c_{4} > 0.

So,

\frac{c_{4}e^{2s}}{(1 + c_{3})(l − δ_{1})} < s^{2} < \frac{c_{3}(2 − 3s)}{l − δ_{1}}.  (4.37)

When l = 1, from the right inequality of (4.37), we have 0 < −s < c_{7}. Therefore, from the left inequality of (4.37), we have −s > c_{8} > 0. That is

0 < c_{8} < −s < c_{7}.

So, from (4.27) and last inequality, we can get c_{3} = 1 + ϵ where ϵ → 0, as N → ∞. Therefore, we can have 0 < δ_{1} < \frac{1}{2} as N is big enough. So, when l = 1, from the right inequality of (4.37), we have

s^{2} + 6s − 4 < 0.

From Lemma 4.4, we know s is an increasing function with respect to l. Therefore,

0 < −s < 3 + \sqrt{13}, for any l < mn/2 ,

when N is big enough. So, from (4.37), we have

\frac{c_{5}}{\sqrt{l}} < −s < \frac{c_{6}}{\sqrt{l}}, as N → ∞.

Because s is an increasing function with respect to k, we have

−s = O(\frac{1}{\sqrt{l}}), as N → ∞. □

If l is fixed we have the following Lemma. Lemma 4.8 If l is fixed and p → p_{0} as N → ∞, we have

s → s(l) as N → ∞,

where s(l) < 0 is only dependent on l. Proof: From Lemma 4.7 we know −s > c > 0. Also,

\frac{1 − e^{ms}}{1 − e^{ns}} ⟶ 1 as N → ∞.

So, from Lemma 4.6, we have

t = −ms − δ_{0}s + log(q/p),

where 0 ≤ δ_{0} ≤ 1 as N → ∞. Therefore, from the second formula of (4.31), we have

∑_{j=1}^{n}(\frac{e^{js−δ_{0}s}}{1 + e^{js−δ_{0}s}} − \frac{e^{js−(1−δ_{0})s}}{1 + e^{js−(1−δ_{0})s}}) = ∑_{j=n+1}^{m} \frac{e^{js−(1−δ_{0})s}}{1 + e^{js−(1−δ_{0})s}} > 0.

Therefore,

\frac{e^{js−δ_{0}s}}{1 + e^{js−δ_{0}s}} − \frac{e^{js−(1−δ_{0})s}}{1 + e^{js−(1−δ_{0})s}} > 0 for all j.

So δ_{0} > 1/2. Because −s > c > 0 we have

∑_{j=1}^{n}(\frac{e^{js−δ_{0}s}}{1 + e^{js−δ_{0}s}} − \frac{e^{js−(1−δ_{0})s}}{1 + e^{(js−(1−δ_{0})s}}) = ∑_{j=n+1}^{m} \frac{e^{js−(1−δ_{0})s}}{1 + e^{js−(1−δ_{0})s}} = e^{ns} ∑_{j=1}^{m−n} \frac{e^{js−(1−δ_{0})s}}{1 + e^{js−(1−δ_{0})s}} = d_{0}e^{ns},

(4.38)

where d_{0} > 0. Because

\frac{e^{js−δ_{0}s}}{1 + e^{js−δ_{0}s}} − \frac{e^{js−(1−δ_{0})s}}{1 + e^{js−(1−δ_{0})s}} = \frac{e^{(j−δ_{0})s} − e^{js−(1−δ_{0})s}}{(1 + e^{(1−δ_{0})s})(1 + e^{δ_{0}s})}

and

e^{(j−δ_{0})s} − e^{js−(1−δ_{0})s} = e^{js} ∑_{u=1}^{∞}((−δ_{0})^{u} − (−(1 − δ_{0}))^{u})\frac{s^{u}}{u!}
= e^{js}(1 − 2δ_{0}) ∑_{u=2}^{∞}((−δ_{0})^{u−1} + … + (−(1 − δ_{0}))^{u−1})\frac{s^{u}}{u!}
= −c_{1}(1 − 2δ_{0})e^{js},

where c_{1} > 0. Therefore, from (4.38), we have

0 < −(1 − 2δ_{0}) < d_{1}e^{ns},

where d_{1} > 0. So

δ_{0} = 1/2 + d_{2}e^{ns},  (4.39)

where d_{2} > 0. From (4.32), we have

∑_{j=1}^{m−1} \frac{je^{js+δ_{0}s}}{1 + e^{js+δ_{0}s}} + ∑_{j=1}^{n} \frac{je^{js−δ_{0}s}}{1 + e^{js−δ_{0}s}} = l.

Now,

∑_{j=1}^{m−1} \frac{je^{js+δ_{0}s}}{1 + e^{js+δ_{0}s}} − ∑_{j=1}^{m−1} \frac{je^{js+ \frac{1}{2} s}}{1 + e^{js+ \frac{1}{2} s}} = ∑_{j=1}^{m−1} \frac{je^{js+ \frac{1}{2} s}(e^{d_{2}e^{ns}} − 1)}{(1 + e^{js+δ_{0}s})(1 + e^{js+ \frac{1}{2} s})}
= d_{2}e^{ns+ \frac{1}{2} s} ∑_{j=1}^{m−1} \frac{je^{js}}{(1 + e^{js+δ_{0}s})(1 + e^{js+ \frac{1}{2} s})}(1 + O(e^{ns}))

(4.40)

In the same way we have

∑_{j=1}^{n} \frac{je^{js−δ_{0}s}}{1 + e^{js−δ_{0}s}} − ∑_{j=1}^{n} \frac{je^{js− \frac{1}{2} s}}{1 + e^{js− \frac{1}{2} s}} = ∑_{j=1}^{n} \frac{je^{js− \frac{1}{2} s}(e^{−d_{2}e^{ns}} − 1)}{(1 + e^{js−δ_{0}s})(1 + e^{js− \frac{1}{2} s})}

= −d_{2}e^{ns+ \frac{1}{2} s} ∑_{j=1}^{m−1}(\frac{je^{js}}{(1 + e^{js+δ_{0}s})(1 + e^{js+ \frac{1}{2} s})} \frac{(1 + e^{js+δ_{0}s})(1 + e^{js+ \frac{1}{2} s})}{(1 + e^{js−δ_{0}s})(e^{s} + e^{js+ \frac{1}{2} s})})(1 + O(e^{ns})).

(4.41)

We know

1 − \frac{(1 + e^{js+δ_{0}s})(1 + e^{js+ \frac{1}{2} s})}{(1 + e^{js−δ_{0}s})(e^{s} + e^{js+ \frac{1}{2} s})} = \frac{(e^{s} − 1) + e^{2js+ \frac{1}{2} s+δ_{0}s}(e^{−2δ_{0}s} − 1) + e^{js}(e^{(1−δ_{0})s} − e^{δ_{0}s})}{(1 + e^{js−δ_{0}s})(e^{s} + e^{js+ \frac{1}{2} s})}
= \frac{(e^{s} − 1) + e^{2js+s}(e^{−s} − 1) + c_{11}e^{js}e^{ns}}{(1 + e^{js−δ_{0}s})(e^{s} + e^{js+ \frac{1}{2} s})}
= \frac{(1 − e^{s})(e^{2js} − 1)}{(1 + e^{js−δ_{0}s})(e^{s} + e^{js+ \frac{1}{2} s})}(1 + e^{ns}) < 0,

where c_{11} > 0. Therefore, from (4.40) + (4.41), we get

∑_{j=1}^{m−1} \frac{je^{js+ \frac{1}{2} s}}{1 + e^{js+ \frac{1}{2} s}} + ∑_{j=1}^{n} \frac{je^{js− \frac{1}{2} s}}{1 + e^{js− \frac{1}{2} s}} = l + c_{12}d_{2}e^{ns}(1 + O(e^{ns})),  (4.42)

where c_{12} > 0. When m_{1} > m and n_{1} > n, we have

∑_{j=1}^{m_{1}−1} \frac{je^{js_{1}+ \frac{1}{2} s_{1}}}{1 + e^{js_{1}+ \frac{1}{2} s_{1}}} + ∑_{j=1}^{n_{1}} \frac{je^{js_{1}− \frac{1}{2} s_{1}}}{1 + e^{js_{1}− \frac{1}{2} s_{1}}} = l + c_{13}d_{2}e^{n_{1}s_{1}}(1 + O(e^{n_{1}s})),  (4.43)

where c_{13} > 0. Subtracting (4.42) from (4.43), we get

∑_{j=1}^{m−1}(\frac{je^{js_{1}+ \frac{1}{2} s_{1}}}{1 + e^{js_{1}+ \frac{1}{2} s_{1}}} − \frac{je^{js+ \frac{1}{2} s}}{1 + e^{js+ \frac{1}{2} s}}) + ∑_{j=1}^{n}(\frac{je^{js_{1}− \frac{1}{2} s_{1}}}{1 + e^{js_{1}− \frac{1}{2} s_{1}}} − \frac{je^{js− \frac{1}{2} s}}{1 + e^{js− \frac{1}{2} s}})

= − ∑_{j=m}^{m_{1}−1} \frac{je^{js_{1}+ \frac{1}{2} s_{1}}}{1 + e^{js_{1}+ \frac{1}{2} s_{1}}} − ∑_{j=n+1}^{n_{1}} \frac{je^{js_{1}− \frac{1}{2} s_{1}}}{1 + e^{js_{1}− \frac{1}{2} s_{1}}} + c_{13}d_{2}e^{n_{1}s_{1}}(1 + O(e^{n_{1}s_{1}})) − c_{12}d_{2}e^{ns}(1 + O(e^{ns}))
= −d_{13}me^{ms_{1}} − d_{14}ne^{ns_{1}} + c_{13}d_{2}e^{n_{1}s_{1}}(1 + O(e^{n_{1}s_{1}})) − c_{12}d_{2}e^{ns}(1 + O(e^{ns})) < 0,

where d_{13} > 0 and d_{14} > 0. Therefore,

0 > c > s > s_{1} as N is big enough.

Because p → p_{0} and s is absolutely bounded when l is fixed, the proof of the Lemma is now completed. □

Consider the following 3 cases.

(a) l → ∞, and mn − l → ∞

(b) l = 0 or l = mn,

(c) l is fixed and finite integer or mn − l is fixed and finite integer. In each of these cases we have a different limiting distribution for P_{U_{m,n}=k}. Put K_{ij} = K_{ij}(s, t). We have the following theorem.

Theorem 4.2 Assume S_{1} = ∑_{i=1}^{N} jI_{j}, S_{2} = ∑_{j=1}^{N} I_{j} where

P (I_{j} = 1) = \frac{m}{n + m} = \frac{m}{N} = p, P (I_{j} = 0) = \frac{n}{N} = 1 − p = q, i = 1, …, N.

Put k = m(m + 1)/2 + l. Then, for the two-sample Wilcoxon rank sum statistic U_{m,n}, as N → ∞, we have

(a) for l → ∞ and l ≤ nm/2, we have

P (U_{m,n} = k) = P (S_{1} = k|S_{2} = m) = e^{−NΛ(k/N,m/N)} \begin{matrix}√ & \overline{pq} \\ √ & \overline{2πN(K_{20}K_{02} − K_{11}^{2})} \end{matrix}(1 + O(l^{−1/2}));

(4.44)

(b) for l = 0,

P (U_{m,n}) = e^{−NΛ(m(m+1)/(2N),m/N)} \frac{1}{P (S_{2} = m)} = p^{m}q^{n}\sqrt{2πNpq}(1 + O(\frac{1}{N}));  (4.45)

(c) for l a fixed and finite integer,

P (U_{m,n} = \frac{m(m + 1)}{2} + l) = e^{−NΛ(k/N,m/N)}f_{s(l)}(l)\sqrt{2πNpq}(1 + O(N^{−1})),  (4.46)

where

f_{s(l)}(l) = \frac{B_{1}(l)e^{ls(l)}}{∏_{j=1}^{∞}(1 + e^{(j− \frac{1}{2} )s(l)}) ∏_{j=1}^{∞}(1 + e^{(j− \frac{1}{2} )s(l)})},

where

B_{1}(l) = ∑_{j=1}^{∞} ♯{i_{1} + i′_{1} + … + i_{j} + i′_{j} − j = l, i_{j_{1}} ≠ i_{j_{2}} and i′_{j_{1}} ≠ i′_{j_{2}} when j_{1} ≠ j_{2} and

i_{j_{1}}, i′_{j_{2}} are positive integers }. Proof: (a) We have

P (S_{1} = k|S_{2} = m) = \frac{P (S_{1} = k, S_{2} = m)}{P (S_{2} = m)} = e^{NK(s,t)−sk−tm} \frac{P (S_{1st} = k, S_{2st} = m)}{P (S_{2} = m)}
= e^{NK(s,t)−sk−tm} \frac{P (S_{1st} = k|S_{2st} = m)P (S_{2st} = m)}{P (S_{2} = m)},

(4.47)

where k = \frac{m(m+1)}{2} + l and

K(s, t) = \frac{1}{N} log Ee^{sS_{1}+tS_{2}} = \frac{1}{N} ∑_{j=1}^{N} log(q + pe^{js+t}).  (4.48)

When l → ∞ and l ≤ nm/2, from Lemma 4.6 and 4.7, we know t = −ms−δs+log(q/p)+ log c_{3} and −s = O(l^{−1/2}). Therefore,

\{ \begin{matrix}K_{10} ∼ N^{−1}l \\ K_{20} = \frac{1}{N} ∑_{j=1}^{N} \frac{pqj^{2}c_{3}e^{js−ms−δs}}{(q+qc_{3}e^{js−ms−δs})^{2}} = N^{2} ∑_{j=0}^{m−1} \frac{c_{3}( \frac{j}{N})^{2}e^{\frac{j}{N} Ns+δs}}{(c_{3}+e^{\frac{j}{N} Ns+δs})^{2}} \frac{1}{N} + N^{2} ∑_{j=1}^{m} \frac{c_{3}( \frac{j}{N})^{2}e^{\frac{j}{N} Ns−δs}}{(1+c_{3}e^{\frac{j}{N} Ns−δs})^{2}} \frac{1}{N} \\ ∼ N^{2} ∫_{0}^{p} y^{2}e^{yNs}dy + N^{2} ∫_{0}^{q} y^{2}e^{yNs}dy ∼ N^{−1}l^{3/2}, \\ K_{30} ∼ N^{−1}l^{2}, \\ K_{40} ∼ N^{−1}l^{5/2}, \\ K_{50} ∼ N^{−1}l^{3} \end{matrix}

(4.49)

Note K_{rv} ∼ K_{r0}, r, v = 1, …, 5. Therefore,

σ_{st}^{2} = K_{20} − \frac{K_{11}^{2}}{K_{02}} ∼ K_{20} ∼ N^{−1}l^{3/2}.

So, Nσ_{st}^{2} ∼ l^{3/2} → ∞. We know

P (S_{1st} = k, S_{2st} = m) = \frac{1}{(2π)^{2}} ∫_{−π}^{π} ∫_{−π}^{π} exp(−iξk − iηm + NK(s + iξ, t + iη) − NK(s, t))dξdη
= J_{1} + J_{2},

where |θ_{1}| < 1, |θ_{2}| < 1 and P_{3}(iξ, iη), P_{4}(iξ, iη), P_{5}(iξ, iη), P_{6}(iξ, iη) are polynomials of degree 3, 4, 5, 6.

J_{1} = ∫_{−ϵ_{2}}^{ϵ_{2}} ∫_{−ϵ_{1}l^{−1/2}}^{ϵ_{1}l^{−1/2}} exp(−iξk − iηm + NK(s + iξ, t + iη) − NK(s, t))dξdη

and

J_{2} = ∫_{(−π,π)^{2}\(−ϵ_{1}l^{−1/2},ϵ_{1}l^{−1/2})×(−ϵ_{2},ϵ_{2})} exp(−iξk − iηm + NK(s + iξ, t + iη) − NK(s, t))dξdη.

Now we look at J_{1} first.

J_{1} = ∫_{−ϵ_{2}}^{ϵ_{2}} ∫_{−ϵ_{1}l^{−1/2}}^{ϵ_{1}l^{−1/2}} exp(−iξk − iηm + NK(s + iξ, t + iη) − NK(s, t))dξdη
= ∫_{−ϵ_{2}}^{ϵ_{2}} ∫_{−ϵ_{1}l^{−1/2}}^{ϵ_{1}l^{−1/2}} exp\{−\frac{1}{2}Nσ_{st}^{2}ξ^{2} − \frac{1}{2}NK_{02}(η + \frac{K_{11}}{K_{02}} ξ)^{2} +

\frac{N}{3!} P_{3}(iξ, iη) + \frac{N}{4!} P_{4}(iξθ_{1}, iηθ_{2})\}dξdη,

where |θ_{1}| < 1, |θ_{2}| < 1 and P_{3}(iξ, iη) and P_{4}(iξ, iη) are polynomials of degree 3 and

4. Put ξ_{1} = \sqrt{N}σ_{st}ξ and η_{1} = \sqrt{NK_{02}}(η + \frac{K_{11}}{K_{02}} ξ). We know K_{rv} ∼ K_{r0} ∼ N^{−1}l^{(r+1)/2}, r, v = 1, 2, 3, 4, l = O(N^{2}) and

\frac{NK_{ij}}{(Nσ_{st}^{2})^{i/2}} ∼ l^{−(i−2)/4}, i = 1, 2, 3, 4.  (4.50)

Therefore,

J_{1} = ∫_{−ϵ_{3}N^{1/2}}^{ϵ_{3}N^{1/2}} ∫_{−ϵ_{4}l^{1/4}}^{ϵ_{5}l^{1/4}} e^{− \frac{1}{2} ξ_{1}− \frac{1}{2} η_{1}}

(1 + P′_{3}(ξ_{1}, η_{1})l^{−1/4} + (P′_{4}(ξ_{1}, η_{1}) + P′_{6}(ξ_{1}, η_{1}))l^{−1/2}e^{\frac{1}{4} ξ_{1}+ \frac{1}{4} η_{1}})dξ_{1}dη_{1} + O(e^{−cN^{1/2}}), where P′_{3}(iξ, iη), P′_{4}(iξ, iη) and P′_{6}(iξ, iη) are polynomials of degree 3, 4 and 6, ϵ_{3} > 0, ϵ_{4} > 0 and ϵ_{5} > 0 can be small enough by chosing ϵ_{1} and ϵ_{2}. Therefore,

J_{1} = \frac{1}{2πN\sqrt{K_{02}}σ_{st}} (1 + O(\frac{1}{l^{1/2}} )) = \frac{1}{2πN\sqrt{K_{20}K_{02} − K_{11}^{2}}} (1 + O(\frac{1}{l^{1/2}} )).

Now let us look at J_{2}. We use a proof related to one used in Albers, Bickel and van Zwet (1976). J_{2} is rewritten as

J_{2} = ∫_{(−π,π)^{2}\(−ϵ_{1}l^{−1/2},ϵ_{1}l^{−1/2})×−ϵ_{2},ϵ_{2}} exp(−iξk − iηm) ∑_{j=1}^{N} \frac{q + pe^{js+t+i(jξ+η)}}{q + pe^{js+t}} dξdη.

When ϵ_{1}l^{−1/2} ≤ |ξ| ≤ π and ϵ_{2} ≤ |η| ≤ π, note −s = O(l^{−1/2}) and t = −ms − δs, where 0 ≤ δ ≤ 1, we have

|∏_{j=1}^{N} \frac{q + pe^{js+t+i(jξ+η)}}{q + pe^{js+t}} | = ∏_{j=1}^{N} \frac{((1 + e^{js−ms−δs})^{2} − 2e^{js−ms−δs}(1 − cos(jξ + η)))^{1/2}}{(1 + e^{js−ms−δs})}
= ∏_{j=1}^{N} (1 − 4\frac{e^{js−ms−δs}}{(1 + e^{js−ms−δs})^{2}} sin^{2}((jξ + η)/2))^{1/2}

≤ exp(−c_{1} ∑_{j=m}^{m+[cl^{1/2}]} sin^{2}((jξ + η)/2)),

where c, c_{1} > 0 and cl^{1/2} ≤ n. For ϵ_{1}l^{−1/2} ≤ |ξ| ≤ ϵ_{1}, we can have |ξ| = ϵ_{1}l_{1}^{−1/2} where 1 ≤ l_{1} ≤ l. Separating the integers m, …, m + [cl^{1/2}] to m, …, m + [cl_{1}^{1/2}]; m + [cl_{1}^{1/2}] + 1, …, m + 2[cl_{1}^{1/2}] and so on, we can get [[cl^{1/2}]/[cl_{1}^{1/2}]] groups. In at least half of the groups at least one third of the j’s satisfy

|jξ + η − 2kπ| > δ > 0 for all k

Therefore,

exp(−c_{1} ∑_{j=m}^{m+[cl^{1/2}]} sin^{2}((jξ + η)/2)) ≤ exp(−\frac{c_{1}[cl^{1/2}]}{6}sin^{2}(ϵ/4)).

For ϵ_{1} ≤ |ξ| ≤ π and ϵ_{2} ≤ |η| ≤ π, we can find l_{3} ≥ [l^{1/2}/2] values of j such that for each of these

|jξ + η − 2kπ| > δ for all k

since in this case two consecutive j’s cannot both be within δ of multiples of 2π. So,

∑_{j=m}^{m+[cl^{1/2}]} c_{1}sin^{2}((jξ + η)/2) ≥ l^{1/2}δ_{1} for all ϵ_{1}l^{−1/2} ≤ |ξ| ≤ π and ϵ_{2} ≤ |η| ≤ π,

where δ_{1} > 0. Therefore,

|∏_{j=1}^{N} \frac{1 + e^{js+t+i(jξ+η)}}{1 + e^{js+t}} | ≤ e^{−δ_{1}l^{1/2}}.

From (4.18), we know

J_{2} = O(\frac{1}{l^{3/2}} ).

Therefore,

P (S_{1st} = k, S_{2st} = m) = \frac{1}{2πN\sqrt{K_{20}K_{02} − K_{11}^{2}}} (1 + O(\frac{1}{l^{1/2}} )).

We know

P (S = m) = \frac{1}{\sqrt{2πNpq}} (1 + O(\frac{1}{N})).  (4.51)

So,

P (U_{m,n} = k) = P (S_{1} = k|S_{2} = m) = e^{−NΛ(s,t)} \begin{matrix}√ & \overline{pq} \\ √ & \overline{2πN(K_{20}K_{02} − K_{11}^{2})} \end{matrix}(1 + O(l^{−1/2})).

(b) We have s → −∞, ms + t → ∞ and (m + 1)s + t → −∞. So, from (4.48), we have

exp(−N lim_{x→0} Λ(k/N, m/N)) = lim_{s→−∞} ∏_{j=1}^{N}(q + pe^{js+t})e^{−mt− \frac{m(m+1)}{2} s}
= lim_{s→−∞} ∏_{j=1}^{m}(qe^{−js−t} + p) ∏_{j=m+1}^{N}(q + pe^{js+t}) = p^{m}q^{n}.
P_{st}(S_{1} = m(m + 1)/2, S_{2} = m) = lim_{s→−∞} \frac{p^{m}q^{n}e^{mt+ \frac{m(m+1)}{2} s}}{∏_{j=1}^{N}(q + pe^{js+t})}
= lim_{s→−∞} \frac{p^{m}q^{n}e^{mt+ \frac{m(m+1)}{2} s}}{p^{m}q^{n}e^{mt+ \frac{m(m+1)}{2} s}} = 1.

Therefore, (4.45) follows from (4.51), (4.21) and (4.24).

(c) When l is fixed, note t = −(m + δ)s + log(q/p), where δ = 1/2 + O(e^{ns}), we have

e^{−NΛ(k/N,m/N)} = ∏_{j=1}^{N}(q + pe^{js+t})e^{−mt−(m(m+1)/2+l)s} = ∏_{j=1}^{m}(qe^{−t−js} + p) ∏_{j=m+1}^{N}(q + pe^{js+t})e^{−ls}
= p^{m}q^{n} ∏_{j=0}^{m−1}(1 + e^{(j+δ)s}) ∏_{j=1}^{n}(1 + e^{(j−δ)s})e^{−ls}
= p^{m}q^{n} ∏_{j=1}^{m}(1 + e^{(j− \frac{1}{2} )s}) ∏_{j=1}^{n}(1 + e^{(j− \frac{1}{2} )s})e^{−ls}(1 + O(e^{ns})).

(4.52)

From Lemma 4.1, s → s(l) < 0 as N → ∞. We can prove s = s(l)(1 + O(ne^{ns(l)})) for some c > 0. Let B_{N1}(l) = ∑_{r=1}^{l} ♯{i_{1} + i′_{1} + … + i_{r} + i′_{r} − r = l, i_{j_{1}} ≠ i_{j_{2}} and i′_{j_{1}} ≠ i′_{j_{2}} when j_{1} ≠ j_{2} and i_{j_{1}} ≤ m, i′_{j_{2}} ≤ n are positive integers } and B_{1}(l) = ∑_{r=1}^{∞} ♯{i_{1} + i′_{1} + … + i_{r} + i′_{r} − r = l, i_{j_{1}} ≠ i_{j_{2}} and i′_{j_{1}} ≠ i′_{j_{2}} when j_{1} ≠ j_{2} and i_{j_{1}}, i′_{j_{2}} are positive integers }. From P (I_{jst} = 1) = pe^{js+t}/(q + pe^{js+t}), we have

P_{st}(S_{1} = m(m + 1)/2 + l, S_{2} = m) = \frac{p^{m}q^{n}B_{N1}(l)e^{mt+(m(m+1)/2+l)s}}{∏_{j=1}^{N}(q + pe^{js+t})}

= \frac{B_{N1}(l)e^{ls}}{∏_{j=0}^{m−1}(1 + e^{(j+δ)s}) ∏_{j=1}^{n}(1 + e^{(j−δ)s})}

= \frac{B_{1}(l)e^{ls(l)}}{∏_{j=1}^{∞}(1 + e^{(j− \frac{1}{2} )s(l)}) ∏_{j=1}^{∞}(1 + e^{(j− \frac{1}{2} )s(l)})}(1 + O(ne^{ns(l)})) = f_{s(l)}(l)(1 + O(ne^{ns(l)})),

(4.53)

where s(l) and f_{s(l)}(l) are only dependent on l. So, (4.46) follows from (4.51), (4.21) and (4.24). □

Remark: Table 2 gives values of B_{1}(l), s(l), exact values of P (U_{m,n} = m(m + 1)/2 + l) and approximations of P (U_{m,n} = m(m + 1)/2 + l) based on (4.46) and (4.44) for m = n = 8. From these we see that the new approximation even for l = 8 is good, however, the normal approximations for k ≤ 4 are very poor.

Table 2 Approximations for two sample Wilcoxon for n = m = 8

k  1  2  3  4  5  6  7  8  9  10
B_{10}(k) = B(k) 1  1  2  2  3  4  5  6  8  10
t(k)  -.889 -.635 -.520 -.451 -.404 -.369 -.342 -.320 -.302 -.286
exact  .0010 .0010 .0020 .0020 .0029 .0039 .0049 .0059 .0078 .0098
P_{new}(W = k) .0010 .0010 .0019 .0019 .0028 .0037 .0044 .0051 .0065 .0077
P_{nor}(W = k) .0012 .0015 .0020 .0026 .0034 .0043 .0054 .0067 .0082 .0098

References

Albers, W., Bickel, P. J. and van Zwet, W. R. (1976), Asymptotic expansions for the Power

of Distribution Free Tests in the One-sample Problem, Ann. Statist. 4 108–156. Daniels, H. E. (1954), Saddlepoint Approximations in statistics, Ann. Math. Statist. 25

631–649. Jensen, J. L. (1995), Saddlepoint Approximations, Oxford Uni. Press Inc. Robinson, J. (1982), Saddlepoint Approximations for Permutation Tests , J. Roy. Statist.

Soc. Ser. B 44 91–101.