The equation for the maximum likelihood estimator for the binomial distribution is $$\mathcal{L}(p|n,y)=\binom{n}{y}p^y(1-p)^{n-y}$$ For a statistical problems I am working on, I am interested in the behaviour of this function only for $p=0.5$. When we set $p=0.5$, we get $$\mathcal{L}(0.5|n,y)=\binom{n}{y}\frac{1}{2^n}$$ My expectation would be, for $y = n/2$, $$\lim_{n\to\infty}\mathcal{L}(0.5|n,n/2)=1$$ However, when I then tried approximating $\mathcal{L}(0.5|n,n/2)$ using Stirling's approximation, I get the following derivation: $$\mathcal{L}(0.5|n,n/2)\approx\sqrt{\frac{2}{\pi n}}$$ which tends to $0$ as $n$ tends to infinity.
My question is, is my derivation wrong, or:
- Does $\lim_{n\to\infty}\mathcal{L}(0.5|n,n/2)=0$,
- Can the Sterling approximation not be applied in this context, or
- Is there some other explanation?
EDIT: After generating some data, I think that my intuition about the limit of this function is wrong, and my derivation is correct. If that's the case - why should I expect $\lim_{n\to\infty}\mathcal{L}(0.5|n,n/2)=0$?