Suppose that X has the Binomial distribution with parameters n , p . How can I show that if ( n + 1 ) p is integer then X has two mode that is ( n + 1 ) p or ( n + 1 ) p − 1 ??

Question

Suppose that   X has the Binomial distribution with parameters   n  ,  p . How can I show that if   (  n  +  1  )  p is integer then   X has two mode that is   (  n  +  1  )  p or   (  n  +  1  )  p  −  1  ??

impire7vw · Accepted Answer

Let       a    k    =  P  (  X  =  k  ), we have      a    k    =                (              n      k                  )            p    k        q          n      −      k          and        a          k      +      1        =                (              n              k        +        1                        )            p          k      +      1            q          n      −      k      −      1        ,where as usual   q  =  1  −  p in binomial distribution.We calculate the ratio                     a                  k          +          1                            a        k            . Note that                           (                    n                  k          +          1                                  )                                  (                    n        k                          )             simplifies to             n      −      k              k      +      1        ,, and therefore            a              k        +        1                    a      k        =            n      −      k              k      +      1        ⋅      p    q    =            n      −      k              k      +      1        ⋅      p          1      −      p        .From this equation we can follow:                    k        &amp;gt;        (        n        +        1        )        p        −        1                ⟹                          a                      k            +            1                          &amp;lt;                  a          k                                    k        =        (        n        +        1        )        p        −        1                ⟹                          a                      k            +            1                          =                  a          k                                    k        &amp;lt;        (        n        +        1        )        p        −        1                ⟹                          a                      k            +            1                          &amp;gt;                  a          k                    The calculation (almost) says that we have equality of two consecutive probabilities precisely if       a          k      +      1        =      a    k  , that is, if   k  =  n  p  +  p  −  1. Note that   k  =  n  p  +  p  −  1 implies that   n  p  +  p  −  1 is an integer.So if   k  =  n  p  +  p  −  1 is not an integer, there is a single mode; and if   k  =  n  p  +  p  −  1 is an integer, there are two modes, at   n  p  +  p  −  1 and at   n  p  +  p.We have been a little casual in our algebra. We have not paid attention to whether we might be multiplying or dividing by   0. We also have casually accepted what the algebra seems to say, without doing a reality check.Suppose that   p  =  0. Then   n  p  +  p  −  1 is an integer, namely   −  1. But whatever   n is, there is a single mode, namely   k  =  0. In all other situations where   n  p  +  p  −  1 is an integer, the   k we have identified is non-negative.However, suppose that   p  =  1. Again,   n  p  +  p  −  1 is an integer, and again there is no double mode. The largest       a    k   occurs at one place only, namely   k  =  n, since   n  p  +  p is in this case beyond our range.That completes the analysis when   n  p  +  p  −  1 is an integer. When it is not, the analysis is simple. There is a single mode, at   ⌊  n  p  +  p  ⌋.

Rey Mcmillan · Accepted Answer

Compute the ratio   b  (  n  ,  p  ;  k  +  1  )  /  b  (  n  ,  p  ;  k  ) and check that this ratio is   &amp;gt;  1 for every   k  &amp;lt;      k    ∗   and   ⩽  1 for every   k  ⩾      k    ∗  , for some integer       k    ∗  .

Suppose that X has the Binomial distribution with parameters n , p . How can I sho

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