534 ANSWERS TO EXERCISES 3.3.1 17. (a) Change of variable, x3 + x3 + t. (b) Induction on n; by definition, z Pno(x-tt)= P(n–1)o(xn -t) dxn. s n (c) The left-hand side is s, dx, . ..~~. dxk+l times lk dxklz* dxk–1 . ..lzz dxl. (d) From (b), (c) we have (r -t) (x + t -ry–1 (x + Pdx) = c 7 (n -r)! t -n). O j, then i/n -F(Xi) > j/n -F(X,). (b) Start with ak = 1.0, bk = 0.0, and ck = 0 for 0 5 k < m. Then do the following for each observation Xj: Set Y + F(X,), k + LmY], ak +-min(ak,Y), bk +-maX(bk,Y), ck + ck + 1. (Assume that F(X,) < 1 so that k < m.) Then set j + 0, rt + r-+-0, and for k = 0, 1, , m -1 (in this order) do the following whenever ck > 0: Set r-+ max(r-, ak -j/n), j + j + ck, r+ + max(r+, j/n-bk). Finally set KL + fir+, K; + fir-. The time required is O(m + n), and the precise value of n need not be known in advance. (If the estimate (k + 4)/m is used for ak and bk, so that only the values ck are actually computed for each k, we obtain estimates of K$ and K; good to within i&i/m, even when m < n.) [ACM Trans. Math. Software 3 (1977), 60-64.1 SECTION 3.3.2 1. The observations for a chi-square test must be independent, and in the second sequence successive observations are manifestly dependent, since the second component, of one equals the first component of the next. 2. Form t-tuples (YJt,. . ,YJtft-l), for 0 5 j < n, and count how many of these equal each possible value. Apply the chi-square test with k = & and with probability l/dt in each category. The number of observations, n, should be at least 5dt.
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