Mentioned that the mode of negative binomial distribution can be found by: t = 1 + ((r-1)/p) where t

britesoulusjhq

britesoulusjhq

Answered question

2022-05-13

Mentioned that the mode of negative binomial distribution can be found by: t = 1 + ((r-1)/p) where t is some number, r is the number of successes and p is the probability of success If t is an integer, there will be 2 modes at t and t-1 If t is not an integer, the integer part of t is the mode.
This is what I have worked on so far: ((r-1)/p) is the expected number of attempts to achieve all required successes excluding the final success The expected number of attempts + the definite final success = the total number of expected attempts.The total number of expected attempts is the mode.
Why could there be two modes of t-1 and t?
P ( X = x ) = ( ( x 1 ) c h o o s e ( r 1 ) ) ( p r ) ( q ( x r ) )
( ( t 2 ) c h o o s e ( r 1 ) ) ( p r ) ( q ( t r 1 ) ) = ( ( t 1 ) c h o o s e ( r 1 ) ) ( p r ) ( q ( t r ) )
( ( t 2 ) c h o o s e ( r 1 ) ) ( q 1 ) = ( ( t 1 ) c h o o s e ( r 1 ) )
( ( t 2 ) ( t 3 ) . . . ( t r ) ) / ( r 1 ) ! x ( q 1 ) = ( ( t 1 ) ( t 2 ) . . . ( t r + 1 ) ) / ( r 1 ) !
( t r ) / q = t 1 [I am stuck here]

Answer & Explanation

hi3c4a2nvrgzb

hi3c4a2nvrgzb

Beginner2022-05-14Added 15 answers

If you have t r q = t 1 then you also have t r = q t q and thus t = r q 1 q = 1 + r 1 p
This says that
- if x < 1 + r 1 p then P ( X = x ) < P ( X = x 1 )
- if x = 1 + r 1 p then P ( X = x ) = P ( X = x 1 )
- if x > 1 + r 1 p then P ( X = x ) < P ( X = x 1 )
So with 1 + r 1 p integer then it and r 1 p have the same highest probabilities and are the modes
while if 1 + r 1 p is not an integer then the mode is the integer immediately below it

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