Show without using the preceding results * that the probability pm(r,n)=n^āˆ’1Em(r,n) of finding exactly š‘š cells empty satisfies pm(r+1,n)=pm(r,n)(nāˆ’m)/n+pm+1(r,n)(m+1)/n (1)

Taylor Barron 2022-11-20 Answered
Show without using the preceding results * that the probability p m ( r , n ) = n āˆ’ 1 E m ( r , n ) of finding exactly š‘š cells empty satisfies
p m ( r + 1 , n ) = p m ( r , n ) n āˆ’ m n + p m + 1 ( r , n ) m + 1 n ( 1 )
* The results which I can't use must be
E m ( r , n ) = ( n m ) A ( r , n āˆ’ m ) = ( n m ) āˆ‘ Ī½ = 0 n āˆ’ m ( āˆ’ 1 ) Ī½ ( n āˆ’ m Ī½ ) ( n āˆ’ m āˆ’ Ī½ ) r
and by association
A ( r , n + 1 ) = āˆ‘ k = 1 r ( r k ) A ( r āˆ’ k , n )
By the way, A ( r , n ) is equal to n ! S ( r , n ) where S ( r , n ) are the Stirling numbers of the second kind.
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Answers (1)

sliceu4i
Answered 2022-11-21 Author has 16 answers
So p m ( r , n ) is the probability of finding m cells out of n empty when you scatter r objects randomly among the bins. Now you are trying to find an expression for m. Think about how you can get there with r + 1 objects. You can either have m cells empty with r objects and put the new one in an occupied bin, or you can have m + 1 cells empty with r objects and put the new one in an unoccupied bin. This should lead you to the equation you want.
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