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International Journal of Scientific and Research Publications, Volume 3, Issue 2, February 2013 648
ISSN 2250-3153
Similarly
Hence it follows from above that
Hence as
Hence the proof.
Lemma 2.5
Out side a set of measure at most the event occurs at least for values of m. i.e out side a set of measure .
.
Proof of the lemma
occurs with positive probability. (by Lemma 2.4)
Suppose
Then there exist an absolute constant , such that .
Let , where summation means that the summation is taken over all’s pairs.
Applying Chebyshev’s inequality, we have for
, there fore outside a set of measure
, there fore, it follows that out side a set of measure , occurs i.e either
Proof of the theorem
III. MAIN RESULTS
Define
We have
and
only if
which implies the occurrence of one of the events.
Hence for
except for a set of measure .
From
, it follows that
After some calculation
It follows that
, we have
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