I saw the following claim in some book without a proof and couldn't prove it myself. d/(dp)P(Bin(n,p)<=d)=−n⋅P(Bin(n−1,p)=d)

acapotivigl 2022-09-14 Answered
I saw the following claim in some book without a proof and couldn't prove it myself.
d d p P ( Bin ( n , p ) d ) = n P ( Bin ( n 1 , p ) = d )
So far I got:
d d p P ( Bin ( n , p ) d ) = d d p i = 0 d ( n i ) p i ( 1 p ) n i = n ( 1 p ) n 1 + i = 1 d ( n i ) [ i p i 1 ( 1 p ) n i p i ( n i ) ( 1 p ) n i 1 ]
But I am not very good playing with binomial coefficients and don't know how to proceed.
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Answers (1)

Jazmin Bryan
Answered 2022-09-15 Author has 12 answers
Step 1
Consider the derivative of the logarithm:
d d p [ log Pr [ X = x p ] ] = d d p [ x log p + ( n x ) log ( 1 p ) ] = x p n x 1 p ,
hence
d d p [ Pr [ X = x p ] ] = ( n x ) p x ( 1 p ) n x ( x p n x 1 p )
and
d d p [ Pr [ X x p ] ] = k = 0 x ( n k ) p k ( 1 p ) n k ( k p n k 1 p ) = k = 0 x ( n k ) k p k 1 ( 1 p ) n k ( n k ) ( n k ) p k ( 1 p ) n 1 k .
Step 2
But observe that
( n k ) ( n k ) = n ! k ! ( n k 1 ) ! = ( k + 1 ) n ! ( k + 1 ) ! ( n ( k + 1 ) ) ! = ( k + 1 ) ( n k + 1 ) ,
hence the second term can be written
( k + 1 ) ( n k + 1 ) p ( k + 1 ) 1 ( 1 p n ( k + 1 ) ) ,
which is the same as the first term except the index of summation has been shifted by 1. Therefore, the sum is telescoping, leaving
d d p [ Pr [ X x p ] ] = 0 ( n x ) ( n x ) p x ( 1 p ) n 1 x .
Step 3
All that remains is to observe
( n x ) ( n x ) = n ! x ! ( n x 1 ) ! = n ( n 1 ) ! x ! ( n 1 x ) ! = n ( n 1 x ) ,
therefore
d d p [ Pr [ X x p ] ] = n Pr [ X = x p ] ,
where X Binomial ( n 1 , p ), as claimed.

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