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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.

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

1 Answer 1

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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.

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

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