Maximum possible variance

Question

From this biology article, end of page 4, the author talks about a random variable which never takes value outside the range $[0,1]$ ($0$ and $1$ included in the range). He says that the maximum variance that this random variable can take equals to the product expected value of the random variable by the expected value of one minus the random variable. In other words:

$$\max\mathrm{Var}\left(X\right) = \mathrm{E}(X) \cdot \mathrm{E}(1-X)$$

Is it true? Why is it true?

Answer 1

It is true. A simple reason is that $$\mathrm{var}(X)=E(X^2)-m^2$$ where $m=E(X)$ and that, if $X$ is almost surely in $[0,1]$, then $X^2\leqslant X$ almost surely hence $E(X^2)\leqslant E(X)=m$, thus $$\mathrm{var}(X)\leqslant m-m^2=m(1-m). $$ More generally, if $0\leqslant X\leqslant x$ almost surely then $\mathrm{var}(X)\leqslant xm-m^2$.

Answer 2

That is the variance of a Bernoulli random variable, which is confined to take only the values 1 or 0. If you hold the expected value constant then concentrating the probability at the extreme values of the variable will make it more variable than letting it take values that can be closer together

Answer 3

as an illustration consider a distribution in which x takes value 1 with probability m and value 0 with probability 1-m. then the expected value is m and the variance is m-m^2 = m*(1-m)