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246 H. El Barmi and H. Mukerjee<br />

The confidence bands could always be improved by the following consideration.<br />

The 100(1−α)% simultaneous confidence bands, [Li, Ui], for Fi, 1≤i≤k, in the<br />

unrestricted case obey the following probability inequality<br />

P(Fi∈ [Li, Ui] : 1≤i≤k)≥1−α.<br />

Under our model, F1≤ F2≤···≤Fk, this probability is not reduced if we replace<br />

[Li, Ui] by [L∗ i , U ∗ i ], where<br />

L ∗ i = max{Lj : 1≤j≤ i} and U ∗ i = min{Uj : i≤j≤ k},1≤i≤k.<br />

6. Hypotheses testing<br />

Let H0 : F1 = F2 =··· = Fk and Ha : F1≤ F2≤···≤Fk. In this section we<br />

propose an asymptotic test of H0 against Ha− H0. This problem has already been<br />

considered by El Barmi et al. [3] when k = 2, and the test statistic they proposed<br />

is Tn = √ n sup x≥0[ ˆ F2(x)− ˆ F1(x)]. They showed that under H0,<br />

(6.1)<br />

lim<br />

n→∞ P(Tn > t) = 2(1−Φ(t)), t≥0,<br />

where Φ is the standard normal distribution function.<br />

For k > 2, we use an extension of the sequential testing procedure in Hogg [7]<br />

for testing equality of distribution functions based on independent random samples.<br />

For testing H0j : F1 = F2 =··· = Fj against Haj− H0j, where Haj : F1 = F2 =<br />

··· = Fj−1≤ Fj,j = 2, 3, . . . , k, we use the test statistic sup x≥0 Tjn(x) where<br />

Tjn = √ n √ cj[ ˆ Fj− Av[ˆF; 1, j− 1]],<br />

with cj = k(j− 1)/j. We reject H0j for large values Tjn, that may be also written<br />

as<br />

Tjn = √ cj[Zjn− Av(Zn; 1, j− 1)],<br />

where Zn = (Z1n, Z2n, . . . , Zkn) T . By the weak convergence result in (3.1) and the<br />

continuous mapping theorem, (T2n, T3n, . . . , Tkn) T converges weakly to (T2, T3, . . . ,<br />

Tk) T , where<br />

Tj = √ cj[Zj− Av[Z; 1, j− 1]].<br />

A calculation of the covariances shows that the Tj’s are independent. Also note<br />

that<br />

d<br />

= Bj(F), 2≤j≤ k,<br />

Tj<br />

where the Bj’s are independent standard Brownian motions and F = �k i=1 Fi =<br />

kF1 under H0. We define our test statistic for the overall test of H0 against Ha−H0<br />

by<br />

Tn = max Tjn(x).<br />

2≤j≤k sup<br />

x≥0<br />

By the continuous mapping theorem, Tn converges in distribution to T, where<br />

T = max<br />

2≤j≤k sup<br />

x≥0<br />

Tj(x).

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