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By William Feller

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Example text

Though the corresponding dyadic expansions ~epresent the' same point 1. Nevertheless, the nO,tion of zero probability enables us to identify the two sample spaces. Stated in more intuitive terms, neglectinga~e~nt ofprobability zero Jhe random choice of a point X between 0 and 1 can be effected by a sequence of coin tossings; conversely, the result of an infinite cointossing game may be represented by a point x of 0, 1. Every random variable of the coin-tossing game may be represented by a function on 0, 1, etc.

For that purpose we restrict t to the interval 0, 1, and in this interval we define v as the inverse function of F. The event {F(X k ) < t} is then identical with the event {Xk :::; vet)} which has probability . F(v(t) = t. Thus P{Yk ~ t} = t as asserted. •• , Yn are mutually independent, and we denote their empirical distribution by Gn • The argument just used shows also that for fixed t the random variable Gn(t) is identical with F n(v(t)). Since t = F(v(t» this implies that at every point of the sample space :R,r.

For reasons of symmetry, F(x) =! throughout an interval of length i centered at x = 1. We have now found three intervals of total length t + t =! in each of which F assumes a constant value, namely 1, i, or 1· Consequently, F can have no jump exceeding i. There remain four intervals of length 116 each, and in each of them the graph of F differs from the whole graph only by a similarity transformation. Each of the four intervals therefore contains a subinterval of half its length in which F assumes a constant value (namely i, t, j, t, respectively).

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