suppose $X$ has the Binomial distribution with parameters $n,p$ . how can show that if $((n+1)p)$ is integer then $X$ has two mode that is $((n+1)p)$ or $((n+1)p-1)?$
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Let $a_k=P(X=k)$. We have $$a_k=\binom{n}{k}p^kq^{n-k},\qquad\text{and}\qquad \binom{n}{k+1}p^{k+1}q^{n-k- 1},$$ where as usual $q=1-p$. We calculate the ratio $\dfrac{a_{k+1}}{a_k}$. Note that $$\frac{\binom{n}{k+1}}{\binom{n}{k}}$$ simplifies to $$\frac{n-k}{k+1},$$ and therefore $$\frac{a_{k+1}}{a_k}=\frac{n-k}{k+1}\frac{p}{q}=\frac{n-k}{k+1}\frac{p}{1-p}.$$ This is $\ge 1$ if $k \le np+p-1$. Thus if $k< np-p+1$, then $a_{k+1}>a_k$, and if $k>np-p+1$, then $a_{k+1}<a_k$. The calculation (almost) says that we have equality of two consecutive probabilities precisely if $a_{k+1}=a_k$, that is, if $k=np+p-1$. Note that $k=np+p-1$ implies that $np+p-1$ is an integer. So unless $k=np+p-1$, there is a single mode, and if $k=np+p-1$ there are two modes, at $np+p-1$ and at $np+p$. Not quite! 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 $np+p-1$ is an integer, namely $-1$. But whatever $n$ is, there is a single mode, namely $k=0$. In all other situations where $np+p-1$ is an integer, the $k$ we have identified is non-negative. However, suppose that $p=1$. Again, $np+p-1$ is an integer, and again there is no double mode. The largest $a_k$ occurs at one place only, namely $k=n$, since $np+p$ is in this case beyond our range. That completes the analysis when $np+p-1$ is an integer. When it is not, the analysis is simple. There is a single mode, at $\lfloor np+p\rfloor$. |
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