estimation of mode of binomial distribution

Question

I know the mode of a binomial $B(n,p)$ distribution is equal to $k = \lfloor (n+1)p\rfloor$. I am wondering how to estimate the probability at this value, i.e. ${n\choose k} p^k (1-p)^{n-k}$, like what's the order of this value in terms of $n$. Is it exp($-\Omega(n)$)?

Answer 1

The mode behaves asymptotically like $np$.

You can use Stirling's approximation to show that $$\binom{n}{np} \sim 2^{nH(p)} (2 \pi n p(1-p))^{-1/2}$$ where $H(p) := - p \log_2 p - (1-p) \log_2(1-p)$ (see here). So, $$\binom{n}{np} p^{np} (1-p)^{n-np} \sim (2 \pi n p(1-p))^{-1/2} =\Theta(n^{-1/2}).$$

In the case $p=1/2$, the above yields the approximation $\sqrt{\frac{2}{\pi n}}$, which is mentioned on Wikipedia (there it is written as $\sqrt{2/\pi} \cdot 2^{-k}$ where $k = \log_4 n = \log_2 \sqrt{n}$).