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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 $(n+1)p$?

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

$$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$

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 $k>(n+1)p$.

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