For given mean $\mu$ of random variable X in [0,1], what is the probability distribution function $p(X)$ that makes $VAR(X)$ maximum?

Question

Given the conditions

$\int_{0}^{1} p(x)dx=1$, $\int_{0}^{1} xp(x)dx=\mu$ and $p(x)\ge0$ for $\forall x \in [0,1]$,

What probability distribution function $p(x)$ makes $Var(X)$=$\int_{0}^{1} x^{2}p(x)dx - \mu^{2}$ maximum?

Is this can be solved using Lagrange multipliers?

Answer 1

The distribution maximizes $Var(X)$ is: $Pr(X=1)=\mu, Pr(X=0)=1-\mu$. Intuitively, you want to pull the distribution to the sides as much as possible under the constraint of $E(X)=\mu$.

Answer 2

Let us prove that the variance is maximized for the corresponding Bernoulli ("might as well put as much probability mass as possible on the extreme points").

Let $X$ be any random variable supported on $[0,1]$, with expectation $\mathbb{E}[X]=\mu\in[0,1]$. Consider the Bernoulli random variable $Y$ such that $$ Y = \begin{cases} 1 &\text{ with probability } \mu\\ 0&\text{ with probability } 1-\mu\\ \end{cases} $$ Clearly, $\mathbb{E}[Y]=\mu\in[0,1]$, and further $\mathbb{E}[Y^2]=\mu$ as well. Now, $$ \mathbb{E}[X^2] = \int_0^1 x^2 p(x) dx \leq \int_0^1 1\cdot x p(x) dx = \mathbb{E}[X] = \mu $$ so $\mathbb{E}[X^2]\leq \mathbb{E}[Y^2]$. As $\operatorname{Var} X = \mathbb{E}[X^2]-\mu^2$, this means $$ \operatorname{Var} X \leq \operatorname{Var} Y\,. $$