Question:
Let a r.v. $0\leq X \leq 1$. Prove that $V(X) \leq \frac{1}{4}$. When do we have an equality?
Answer:
1- $0\leq X \leq 1 \Rightarrow X^2 \leq X \Rightarrow \int X(\omega)^2 dP (\omega) = E[X^2] \leq E[X] = \int X(\omega) dP (\omega)$.
So $V(X) = E[X^2] - E^2[X] \leq E[X] - E^2[X] = E[X](1-E[X])$
2- The polynomial $x-x^2$ get its maximum on at $x=1/2$ and in such a case has a value of $1/4$. So we can conclude without a doubt at least that $V(X) \leq 1/4$
3- But maybe we can be more precise and write $V(X) < 1/4 $. In order to verify this hypothesis let check if it is possible to have $V(X)=1/4$.
In such a case we will have necessary that $E[X] - E^2[X] = 1/4 \Rightarrow E[X] =1/2 $
4- More over on one side $V(X) + E^2[X] = E[X^2]$, while on an other side we have that $ V(X) + E^2[X] = 1/4 + 1/4 = 1/2 \Rightarrow E[X^2] = 1/2$. Thus $E[X^2] = E[X] \Rightarrow E[X(1-X)] = 0 $. Because $0 \leq X \leq 1 \Rightarrow X(1-X) \geq 0$. So we conclude that $X(1-X) =0 \Rightarrow X =0 $ or $X=1$. Thus $X$ is a Bernoulli r.v. with $E[X]= 1/2 \Rightarrow $ probability of $X=1$ is $1/2$ while the probability of $X=0$ is too $1/2$.
Q.E.D.
Is it correct? I have mostly doubt on my part "4-"
Thank you.
With the help of https://math.stackexchange.com/users/1064504/geetha290krm comment I have slightly improve my answer and now it seems that my solution is correct.
I am just writing this message in order to mark this question as solved.