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.

Question

I saw the following claim in some book without a proof and couldn&#039;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&#039;t know how to proceed.

Jazmin Bryan · Accepted Answer

Step 1Consider 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 2But 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 3All 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.

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)

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