How to show that there is only one possible distribution for $X$ with variance $\frac{1}{4}$, where $X \in [0,1]$.

Question

Let $X$ be an r.v. (discrete or continuous) such that $0 ≤ X ≤ 1$ always holds. Let $μ = \mathbb{E}(X)$. Show that $$a.) \ Var(X) \leq \mu - \mu^2 \leq \frac{1}{4} \\ b.) \ \textrm{ Show that there is only one possible distribution for }X \textrm{ for which } Var(X) = \frac{1}{4}. \\ \textrm{ What is the name of this distribution} ? $$

I have solved part $a.$ $$X-1\leq 0 \\ \implies X(X-1)\leq 0 \\ \implies X^2\leq X \\ \implies \mathbb{E}(X^2) \leq \mathbb{E}(X)=\mu\\ \implies \mathbb{E}(X^2)-\mu^2 \leq \mu-\mu^2 \\ \implies Var(X) \leq \mu-\mu^2$$

Now to show $\mu - \mu^2 \leq \frac{1}{4}$, let us consider $:\\ f(\mu)=\mu-\mu^2\\ \implies f'(\mu)=1-2\mu \\ \textrm{Setting } f'(\mu)=0, \textrm{ we have }\mu=\frac{1}{2}\\ \textrm{Also we have } f''(\mu)|_{\mu=\frac{1}{2}}=-2<0\\ \therefore f(\mu) \textrm{ has the maximum value } \frac{1}{4} \textrm{ at } \mu=\frac{1}{2}\\ \textrm{Hence } f(\mu)=\mu-\mu^2\leq\frac{1}{4} \\ \therefore Var(X) \leq \mu - \mu^2 \leq \frac{1}{4} $

For $b.$ how do I show that there is only one possible distribution for $X$ with variance $\frac{1}{4}$

Any help is valuable and appreciated.

Answer 1

You've already shown that this can only be maximized when $\mu = 1/2$. You need to use this. Another characterization of the variance is that it is the square of the average "distance from the mean". If the mean $\mu$ is 1/2, how far from that can X be at most? How does that compare to the bound?

Answer 2

One has $$E(X-c)^2 = \mathrm{Var}(X)+(E(X)-c)^2,\quad |X-1/2|\leq 1/2$$ Hence $$\mathrm{Var}(X)\leq E(X-1/2)^2 \leq 1/4 $$ The first inequality holds with equality if and only if $E(X) = 1/2$. The second holds with equality if and only if $|X-1/2| = 1/2$ almost surely, where we have used that $E(U) = 0$ for $U\geq 0$ if and only if $U = 0$ almost surely.

Hence $X\in \{0,1\}$ almost surely and $P(X=1)=E(X)=1/2$. We conclude $X\sim \mathrm{Bern}(1/2)$.