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