Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a binomial random variable with parameters $n$ and $p$.

How do I show the following? $$P(X=k+1)=\frac{p}{1-p}\frac{n-k}{k+1}P(X=k), {\ }k=0,1,...,n-1$$ As $k$ goes from $0$ to $n$, $P(X=k)$ first increases and then decreases. How do I show that this probability reaches its largest value when $k$ is the largest integer less than or equal to $(n+1)p$?

share|cite|improve this question
up vote 1 down vote accepted

One way of doing this would be to use the pdf of a binomial:


from which you can figure out what $P(X=k+1)/P(X=k)$ equals.

To see that it's increasing up to $(n+1)p$, notice that for $k\leq (n+1)p-1$

$$\frac{p}{1-p}\frac{n-k}{k+1}\geq \frac{p}{1-p}\frac{n-(n+1)p+1}{(n+1)p}=\frac{p}{1-p}\frac{n(1-p)-p+1}{(n+1)p}=\frac{p(1-p)n+p(1-p)}{p(1-p)n}> 1$$

where I assumed $p\neq 0,1$ (both frivolous cases). So that ratio is strictly greater than 1 for $0<k<(n+1)p$. A very similar argument will show that the ratio is less than 1 for $k>(n+1)p$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.