Step 2



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Step 2

Any distribution supported on with a mean differing by say from the nearest integer must belong to the set of the previous section for . Thus for large enough the set is a subset of for some positive . According to Theorem 1

for all in . In view of the inequality

and 7 above we find

On the explosion set there is a (random) such that the right hand of this inequality is less than for all . A Taylor expansion of the logarithm thus shows

 

for all large and all in . Since almost surely the lemma is proved.



Richard Lockhart
Thu Oct 26 23:26:04 PDT 1995