Show that if $P(0 \leq X \leq c)=1$ then $Var(X) \leq \frac{c^2}{4}$

I need to show that if $$P(0 \leq X \leq c)=1$$ then $$Var(X) \leq \frac{c^2}{4}$$

I can show that using 2 things: First, that $E[X^2] \leq cE[X]$ and secondly that $Var(X) \leq c^2[\alpha(1-\alpha)]$ for $\alpha=\frac{E[X]}{c}$.

Could anyone help me prove these 2 steps?

Last step is quite immediate.

• What does $a$ mean in $Var(X) \leq c^2[\alpha (a - \alpha)]$ pls? Jun 2, 2015 at 12:44
• I am terribly sorry. It should be 1. Edited original question.
– E Be
Jun 2, 2015 at 12:48

First Proof

Define a new random variable $Y = X - \frac{c}{2}$. It is easy to know that $$Var(Y) = Var(X)$$ by using the fact $Var(X+a)=Var(X)$. Thus we only need to show $$Var(Y) \leq \frac{c^2}{4}$$

Since $0 \leq X \leq c$, we have $$-\frac{c}{2} \leq Y \leq \frac{c}{2}$$ thus $$E[Y^2] \leq \frac{c^2}{4}$$

Therefore, $$Var(Y) = E[Y^2] - (E[Y])^2 \leq \frac{c^2}{4} - (E[Y])^2 \leq \frac{c^2}{4}$$

Another Proof

• We first prove that $E[X^2] \leq cE[X]$. This is easy since $E[X^2] \leq E[cX] = cE[X]$.

• We next prove that $Var(X) \leq \frac{c^2}{4}$. Note that $$Var(X) = E[X^2] - E[X]^2 \leq c(E[X] - \frac{E[X]^2}{c}) \tag{1}$$ Moreover, \begin{align} &(2E[X] - c)^2 \\ =\ &4E[x]^2 - 4cE[x] + c^2 \\ =\ &4c(\frac{E[X]^2}{c} - E[x] + \frac{c}{4}) \\ \geq\ &0 \end{align} thus, we have $$E[X] - \frac{E[X]^2}{c} \leq \frac{c}{4} \tag{2}$$

According to (1) and (2), it is proved.

• That's great but I need some approach for those 2 steps I wrote down at the question.
– E Be
Jun 2, 2015 at 12:38
• @UdiBehar I see. Let me try. Jun 2, 2015 at 12:40
• @UdiBehar Is the new proof you want? Jun 2, 2015 at 13:00
• Wonderful. Thanks a lot.
– E Be
Jun 2, 2015 at 13:02