Let Y be a random variable with $0\le Y\le 1.$

Question

Let Y be a random variable with $$0\le Y\le 1.$$Show that $$var(Y)\le 1/4 $$ and that $$var(Y)= 1/4 $$ if and only if P(0)=1/2=P(1).

Answer 1

If $0\le Y\le 1$ with probability $1$, we have $Y^2\le Y$ with probability $1$. Thus the variance of $Y$ is $\le \mu-\mu^2$, where $\mu$ is the mean. Note that the function $t-t^2$ attains a maximum of $\frac{1}{4}$ at $t=\frac{1}{2}$.

The maximum is attained precisely if $Y=Y^2$ with probability $1$, and the mean is $\frac{1}{2}$. Thus with probability $1$ we must have $Y=0$ or $Y=1$. The condition $\mu=\frac{1}{2}$ does the rest. For if $Y=1$ with probability $p\ne \frac{1}{2}$, then $\mu\ne \frac{1}{2}$.

Answer 2

Since $Y(1-Y)\geq0$ we conclude that $$ \mathbb{E}Y-(\mathbb{E}Y)^2-\mathrm{var}(Y)=\mathbb{E}Y-\mathbb{E}Y^2=\mathbb{E}(Y-Y^2)\geq0 $$ Thus $$\mathrm{var}(Y)\leq(\mathbb{E}Y)(1-\mathbb{E}Y)\leq \frac{1}{4}$$ Now equality folds if and only if $\mathbb{E}Y=1/2$ and $Y=Y^2$, $P$-a.s., or equivalently $P(Y\notin\{0,1\})=0$. That is if $Y$ is a Bernoulli trial with equal probability.