Question

Let $X$be a random variable that takes on values between$0$and$c$. That is$P\left\{0\le X\le c\right\}=1$.Show that

$Var\left(X\right)\le \frac{c^2}{4}$

Hint: One approach is to first argue that

localid="1646883602992" $E\left[X^2\right]\le cE\left[X\right]$

and then use this inequality to show that

$Var\left(X\right)\le c^2\left[\alpha \right(1-\alpha \left)\right]where\alpha =\frac{E\left[X\right]}{c}$

Short answer

Hence, proved that$Var\left(X\right)\le \frac{c^2}{4}$.

Step-by-step solution

Given information:

Let$X$be a random variable that takes on values between$0$and$c$. That is,$P\left\{0\le X\le c\right\}=1$

Hint: One approach is to first argue that

$E\left[X^2\right]\le cE\left[X\right]$

and then use this inequality to show that

$Var\left(X\right)\le c^2\left[\alpha \right(1-\alpha \left)\right]where\alpha =\frac{E\left[X\right]}{c}$

Explanation:

Since$X$be, a random variable that takes on values between $0$and$c$, we know that

$f\left(x\right)=0ifx\le 0orx>0$

Thus,

$E\left[X^2\right]={\int }_0^c{x}^{2}f\left(x\right)dx\le {\int }_0^{c}cxf\left(x\right)dx=cE\left[X\right]$

Explanation:

Hence,

$Var\left(X\right)=E\left[X^2\right]-E{\left[X\right]}^2\le cE\left[X\right]-E{\left[X\right]}^2=c^2(\alpha -{\alpha }^2)$

Where$\alpha =\raisebox{1ex}{$E\left[X\right]$}\!\left/ \!\raisebox{-1ex}{$c$}\right.$we know$\alpha -{\alpha }^2$attains its maximum value of$\frac{1}{4}at\alpha =\frac{1}{2}$so

$Var\left(X\right)\le c^2{\overline{)(\alpha -{\alpha }^2)}}_{\alpha =\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=c^2\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{c^2}{4}$