When is the maximum value attained in a binomial distribution ?

$$\begin{aligned} f(k, n, p) = {n \choose k} p^k (1-p)^{n-k} \end{aligned}$$

Your first instinct might consider figuring out the derivative but that would quickly turn into a horrific mess. A nice route is to consider the $k+1$st term against the $k$th term:

$$\begin{aligned} \rho &= {f(k+1, n, p) \over f(k, n, p)} \\ &= {{n \choose k+1} p^{k+1} (1-p)^{n-k-1} \over {n \choose k} p^k (1-p)^{n-k}} \\ &= {(n-k) p \over (k+1)(1-p) } \\ \end{aligned}$$

If $\rho > 1$, $f$ increases as $k$ increases, whereas if $\rho < 1$, $f$ decreases as $k$ increases. The maximum is attained when $\rho = 1$, which is when:

$$\begin{aligned} (k+1)(1-p) &= (n-k) p \\ k+1-p &= np \\ k+1 &= (n+1)p \\ \end{aligned}$$

So if $(n+1)p$ is an integer, that would be the value of $(k+1)$ that would yield the maximum value $M$.

$$\begin{aligned} M = f(k+1, n, p) = f((n+1)p, n, p) \\ \end{aligned}$$

But given $\displaystyle \rho = 1 = {f(k+1, n, p) \over f(k, n, p)}$, both $(k+1)$ and $k$ would yield the same $M$:

$$\begin{aligned} M &= f(k+1, n, p) = f((n+1)p, n, p) \\ &= f(k, n, p) = f((n+1)p-1, n, p) \\ \end{aligned}$$

If $(n+1)p$ is not an integer, then M would be attained when $k + 1 = \lfloor (n+1)p \rfloor$ or $k = \lfloor (n+1)p \rfloor - 1$.

Therefore, $k \approx np$.

$\Box$

# posted by rot13(Unafba Pune) @ 4:54 PM