Maeve Bowers

2022-05-03

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

Beginner2022-05-04Added 19 answers

Let ${a}_{k}=P(X=k)$, we have

${a}_{k}={\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{p}^{k}{q}^{n-k}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{a}_{k+1}={\textstyle (}\genfrac{}{}{0ex}{}{n}{k+1}{\textstyle )}{p}^{k+1}{q}^{n-k-1},$

where as usual $q=1-p$ in binomial distribution.

We calculate the ratio $\frac{{a}_{k+1}}{{a}_{k}}$. Note that $\frac{{\textstyle (}\genfrac{}{}{0ex}{}{n}{k+1}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}}$ simplifies to $\frac{n-k}{k+1},$, and therefore

$\frac{{a}_{k+1}}{{a}_{k}}=\frac{n-k}{k+1}\cdot \frac{p}{q}=\frac{n-k}{k+1}\cdot \frac{p}{1-p}.$

From this equation we can follow:

$\begin{array}{r}k>(n+1)p-1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}_{k+1}<{a}_{k}\\ k=(n+1)p-1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}_{k+1}={a}_{k}\\ k<(n+1)p-1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}_{k+1}>{a}_{k}\end{array}$

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 if $k=np+p-1$ is not an integer, there is a single mode; and if $k=np+p-1$ is an integer, there are two modes, at $np+p-1$ and at $np+p$.

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 $np+p-1$ 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 $.

${a}_{k}={\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{p}^{k}{q}^{n-k}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{a}_{k+1}={\textstyle (}\genfrac{}{}{0ex}{}{n}{k+1}{\textstyle )}{p}^{k+1}{q}^{n-k-1},$

where as usual $q=1-p$ in binomial distribution.

We calculate the ratio $\frac{{a}_{k+1}}{{a}_{k}}$. Note that $\frac{{\textstyle (}\genfrac{}{}{0ex}{}{n}{k+1}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}}$ simplifies to $\frac{n-k}{k+1},$, and therefore

$\frac{{a}_{k+1}}{{a}_{k}}=\frac{n-k}{k+1}\cdot \frac{p}{q}=\frac{n-k}{k+1}\cdot \frac{p}{1-p}.$

From this equation we can follow:

$\begin{array}{r}k>(n+1)p-1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}_{k+1}<{a}_{k}\\ k=(n+1)p-1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}_{k+1}={a}_{k}\\ k<(n+1)p-1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}_{k+1}>{a}_{k}\end{array}$

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 if $k=np+p-1$ is not an integer, there is a single mode; and if $k=np+p-1$ is an integer, there are two modes, at $np+p-1$ and at $np+p$.

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 $np+p-1$ 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 $.

Rey Mcmillan

Beginner2022-05-05Added 11 answers

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