# Finding mode in Binomial distribution

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 $(n+1)p-1?$

• Hint: compute the ratio $b(n,p;k+1)/b(n,p;k)$ and check that this ratio is $\gt1$ for every $k\lt k^*$ and $\leqslant1$ for every $k\geqslant k^*$, for some integer $k^*$.
– Did
Commented Mar 8, 2012 at 16:27

Let $$a_k=P(X=k)$$, we have $$a_k=\binom{n}{k}p^kq^{n-k}\qquad\text{and}\qquad a_{k+1}=\binom{n}{k+1}p^{k+1}q^{n-k- 1},$$ where as usual $$q=1-p$$ in binomial distribution.

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 $$\frac{a_{k+1}}{a_k}=\frac{n-k}{k+1}\cdot\frac{p}{q}=\frac{n-k}{k+1}\cdot\frac{p}{1-p}.$$

From this equation we can follow:

\begin{align} k > (n+1)p-1 \implies a_{k+1} < a_k \\ k = (n+1)p-1 \implies a_{k+1} = a_k \\ k < (n+1)p-1 \implies a_{k+1} > a_k \end{align}

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 $$np+p$$.

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 $$\lfloor np+p\rfloor$$.

To summarize in a single formula: $$\mathrm{mode(s)} = \begin{cases} 0, & \text{if } p = 0; \\ n, & \text{if } p = 1; \\ np + p - 1 \text{ and } np + p, & \text{if } np + p - 1 \in \mathbb{Z} \text{ and } 0 < p < 1; \\ \lfloor np + p \rfloor, & np + p - 1 \notin \mathbb{Z}. \end{cases}$$

• I am solving a similar exercise and I have some doubts: why taking the ratio $\frac{p_X(k+1)}{p_X(k)}$ gives you the mode (most probable value), which is defined as $\sup_{x \in R_X} p_X(x)$? Commented Sep 7, 2014 at 23:26
• The sup is in this case a max, since the random variable takes on integer values. Apart from a couple of "degenerate" cases pointed out in the answer, the probabilities rise and then fall. The ratio of consecutive terms is therefore $\gt 1$ for a while, then $\lt 1$, except that in somewhat unusual cases we can have ratio $1$, so two consecutive values each qualify as a mode. Looking at the ratios tells us when the probability has reached a maximum. Commented Sep 8, 2014 at 1:38
• Thanks very much for the explanation. Commented Sep 8, 2014 at 2:47
• You are welcome. That was an overview. The detail is in the answer above. Commented Sep 8, 2014 at 2:48