558 ANSWERS TO EXERCISES 3.5 16. (Solution by
558 ANSWERS TO EXERCISES 3.5 16. (Solution by R. P. Stanley.) Whenever the subsequence S = (b -l), (b - 2), . . . , 1, 0, 0, 1, . , (b -2), (b -1) appears, a coupon set must end at the right of S, since some coupon set is completed in the first half of S. We now proceed to calculate the probability that a coupon set begins at position n by manipulating the probabilities that the last prior appearance of S ends at position 72 - 1, n -2, etc., as in exercise 15. 18. Proceed as in the proof of Theorem A to calculate fi and Pr. 19. (Solution by T. Herzog.) Yes; e.g., the sequence (Uln,s~) when (Un) satisfies R4 (or even its weaker version), cf. exercise 33. 21. Pr(& E Ml,. . . , i&+&l E Mk) = p(Ml). . .p(Mk), for all MI, , Mk E ht. 22. If the sequence is k-distributed, the limit is zero by integration and Theorem B. Conversely, note that if f(zl, . . , zk) has an absolutely convergent Fourier series f(X1,. . . , xk) = U(Cl, , Ck) eXp(%i(clzl + + ckxk)), c -co