3.3.1 GENERAL TEST PROCEDURES 55 To study the (Web hosting directory)
3.3.1 GENERAL TEST PROCEDURES 55 To study the distribution of K,f and K;, we begin with the following basic fact: If X is a random variable with the continuous distribution F(z), then F(X) is a uniformly distributed real number between 0 and 1. To prove this, we need only verify that if 0 5 y 5 1 we have F(X) 5 y with probability y. Since F is continuous, F(Q) = y for some ~0; thus the probability that F(X) 5 y is the probability that X 5 20. By definition, the latter probability is F(Q), that is, it is y. Let YJ = nF(X,), for 1 2 j 5 n, where the X s have been sorted as in Step 2 above. Then the variables Yy are essentially the same as independent, uniformly distributed random numbers between 0 and 1 that have been sorted into nondecreasing order, Yl 2 YZ 5 . .. 2 Y,; and the first equation of (13) may be transformed into Kz = Lmax(l-Y1,2-Yz ,…, n-Y,). 6 If 0 5 t 5 n, the probability that K,+ 5 t/J;; is therefore the probability that Yj 2 j-t for 1 5 j 5 n. This is not hard to express in terms of n-dimensional integrals, where aj = max(j -t, 0). (24) The denominator here is immediately evaluated: it is found to be nn/n!, which makes sense since the hypercube of all vectors (yl, ~2,. . . , yn) with 0 2 yj < n has volume nn, and it can be divided into n! equal parts corresponding to each possible ordering of the y s. The integral in the numerator is a little more difficult, but it yields to the attack suggested in exercise 17, and we get the general formula probability that (KL 5 $) = ~o~<~(~)(ic-t)*(t+li-ic)ntl- (25) The distribution of K; is exactly the same. Equation (25) was first obtained by Z. W. Birnbaum and Fred H. Tingey [Annals Math. Stat. 22 (1951), 592-5961; it may be used to extend Table 2. In his original paper, Smirnov proved that lim probability that (KL 2 s) = 1 - C2 , if s > 0. (26) 12 00 This together with (25) implies that, for all s 2 0, we have