Supremum of a Cumulative Distribution function

Question

How to find supremum of the following function

1. For $x$ belonging to the entire real line, show that the supremum of $\frac{F(x)(1-F(x))}{n} \leq \frac{1}{4n}$ where $F(x)$ is the cumulative distribution function

Answer 1

What do you mean by the $n$ and the $N$? Are they the same one? If they are equal, then the inequality becomes $$F(x)(1-F(x))\leq \frac{1}{4}.$$ It can be deduced by am-gm inequality:$$F(x)(1-F(x))\leq \left(\frac{F(x)+(1-F(x))}{2}\right)^2=\frac{1}{4}.$$

Answer 2

\begin{align} F(x)(1-F(x)) = p(1-p) & = \overbrace{p - p^2 = -\left( p^2 - p + \frac 1 4 \right) + \frac 1 4 }^{\Large\text{completing the square}} \\[10pt] & = \frac 1 4 - \left( p-\frac 1 2 \right)^2 \quad \begin{cases} = 1/4 & \text{if } p=1/2, \\ < 1/4 & \text{if } p\ne 1/2. \end{cases} \end{align}

If $F$ is everywhere continuous, and being a c.d.f. goes from $0$ up to $1,$ then for some value of $x$ you have $F(x)=1/2,$ so that $F(x)(1-F(x))=1/4.$

But if $F$ has discontinuities, it may jump from some number less than $1/2$ to some number more than $1/2,$ and then the biggest that $F(x)(1-F(x))$ ever gets would be less than $1/4.$