So I'm working on a problem which I've been unable to figure out.
So the problem goes as follows:
Let $X$ be binomially distributed with $0<p<1$. So $p(X=k)=C_{n,k}p^k(1-p)^{n-k}, k=0,1,\ldots ,n$. $p(X=k)=p(k)$ and let $0\leq k_0 \leq n$ be the largest index for which $p(k_0)=max_{0\leq k_0 \leq n}\cdot p(k)$.
Show that $k_0=\left \lfloor{np+p}\right \rfloor$, the largest whole number smaller or equal to $np+p$.
One of my class mates did manage to solve it with a proof that's over 3 pages long, but I've seen no other solutions yet, so I'm really hoping for some insight.
thanks in advance!
We have $p(k) = {n \choose k}p^k(1-p)^{n-k}$. Notice that we have $$ \frac{p(k)}{p(k-1)} = \frac{{n \choose k}p^k(1-p)^{n-k}}{{n \choose {k-1}}p^{k-1}(1-p)^{n-k+1}} = \frac{(k-1)!(n-k+1)!}{k!(n-k)!} \frac{p}{1-p} = \frac{n-k+1}{k} \frac{p}{1-p}.$$ Thus, we have $\frac{p(k)}{p(k-1)} \geq 1$ if and only if $$ \frac{n-k+1}{k} \frac{p}{1-p} \geq 1$$ $$ p(n+1) - kp \geq k(1-p) $$ $$k \le pn + p. $$ In other words, we have that $p(k)$ is increasing in $k$ until it gets to $\lfloor pn + p\rfloor$ and then it starts decreasing. This means its maximum value is attained at $\lfloor pn + p\rfloor$.