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532 ANSWERS TO EXERCISES 3.3.1 shown that for large values of n the function Fn(z) behaves as a Brownian motion, and techniques from this branch of probability theory may be used to study it. The situation is exploited in articles by J. L. Doob and M. D. Donsker, Annals Math. Stat. 20 (1949), 3933403 and 23 (1952), 277-281; this is generally regarded as the most enlightening way to study the KS tests.] 7. ((Cf. Eq. (13).) Take j = n to see that K$ is never negative an:1 it can get as high as fi. Similarly, take j = 1 to make the same observations about K;. 8. The new KS statistic was computed for 20 observations. The distribution of K&, was used as F(z) when the KS statistic was computed. 9. The idea is erroneous, because all of the observations must be independent. There is a relation between the statistics K$ and K; on the same data, so each test should be judged separately. (A high value of one tends to give a low value of the other.) Similarly, the entries in Figs. 2 and 5 (which show 15 tests for each generator) do not show 15 independent observations, because the maximum-of-5 test is not independent of the maximum-of-4 test. The three tests of each horizontal row are independent (because they were done on different parts of the sequence), but the five tests in a column are somewhat correlated. The net effect of this is that the 95-percent probability levels, etc., which apply to one test, cannot legitimately be applied to a whole group of tests on the same data. Moral: When testing a random number generator, we may expect it to pass each of several tests, e.g., the frequency test, maximum test, run test; but an array of data from several different tests should not be considered as a unit since the tests themselves may not be independent. The K$ and K; statistics should be considered as two separate tests; a good source of random numbers will pass both of the tests. 10. Each Y, is doubled, and np, is doubled, so the numerators of (6) are quadrupled while the denominators only double. Hence the new value of V is twice as high as the old one. 11. The empirical distribution function stays the same; the values of Kz and K; are multiplied by fi. 12. Let 2, = (Ys -nqS)/Jnsg. The value of V is n times c (4s -Ps + ~zs)2/p,, l h/2 occurs with very small probability; then IFn(z) -F(z)1 will be improbably high under the assumed distribution F(z). 13. (The max notation should really be replaced by sup since a least upper bound is meant; however, max was used in the text to avoid confusing too many readers by the less familiar sup notation.) For convenience, let Xs = -00, X,+I = +oo. When X, 2 z < Xj+l, we have Fn(z) = j/n; therefore max(F,(z) -F(s)) = j/n -F(X,) and max(F(z) -Fn(z)) = F(X,+r) -j/n in this interval. As j varies
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