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]

Question

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]

hi3c4a2nvrgzb · Accepted Answer

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  &amp;lt;  1  +                    r        −        1            p       then   P  (  X  =  x  )  &amp;lt;  P  (  X  =  x  −  1  )- if   x  =  1  +                    r        −        1            p       then   P  (  X  =  x  )  =  P  (  X  =  x  −  1  )- if   x  &amp;gt;  1  +                    r        −        1            p       then   P  (  X  =  x  )  &amp;lt;  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 modeswhile if   1  +                    r        −        1            p       is not an integer then the mode is the integer immediately below it

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

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