How to prove that maximum occurs at x=n/2 when n is even

Question

I am stuck with this question: Using the recursion formula of a binomial distribution and for $\theta = \frac{1}{2}$:

$b(x+1;n,\theta) = \frac{\theta(n-x)}{(x+1)(1-\theta)}*b(x;n,\theta)$

Show that the binomial distribution has a maximum at $x=\frac{n}{2}$ when $n$ is even; and maxima at $x=\frac{n-1}{2}$ and $x=\frac{n+1}{2}$ when $n$ is odd.

I'm not too sure how to approach the question, can anyone kindly shed some light please? Thank you!

Answer 1

What do you mean by "differentiating" when there are no "differentiation" on $\mathbb{N}$?

Instead the recursion gives the ratio of successive $b$'s $$ \frac{b(x+1;n,\theta)}{b(x;n,\theta)}=\frac{\theta(n-x)}{(x+1)(1-\theta)} $$ so just show it grows larger as you move each step to $x=\frac{n}{2}$ ($n$ even) or $x=\frac{n\pm 1}{2}$ ($n$ odd) when $\theta=\frac{1}{2}$.

Answer 2

The maximum occurs when the ratio crosses the unit threshold.

We solve $$\theta(n-x)=(x+1)(1-\theta)$$

or

$$x=\theta(n+1)-1.$$

For $\theta=\dfrac12$, $\dfrac{n-1}2$.