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A random variable X has range from $(0,a)$. Show that $Var(X)\le a^2/4$.
The given hint is to show $E[X^2]\le aE[X]$ first, and use that to show $Var(X)\le a^2[\beta(1-\beta)]$, where $\beta=E[X]/a$.
I don't know how to start. I've tried using Cheyshev's inequality but it doesn't work. Can anyone give me a hint how to start? Any concept I've been missing?

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  • $\begingroup$ is that all the information in the question? $\endgroup$ – Siong Thye Goh Nov 27 '18 at 1:12
  • $\begingroup$ @SiongThyeGoh I've edited some typos. And yes, that is all $\endgroup$ – Yibei He Nov 27 '18 at 1:19
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$$E[aX-X^2]=E[X(a-X)]\ge 0$$

$$Var(X)=E[X^2]-E[X]^2\le aE[X]-E[X]^2$$

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You know that $Var(X)=\int_0^a(x-\mu)^2dx=\int_0^ax^2p(x)dx-\left(\int_0^axp(x)dx\right)^2$, where $p(x)\geq0$ is the probability distribution and $\mu=E(X)$. Hence, $$ Var(X)\leq a\int_0^axp(x)dx-\mu^2=\mu(a-\mu)=a^2\beta(1-\beta)=-a^2(\beta^2-\beta+1/4-1/4)=-a^2(\beta-1/2)^2+a^2/4\leq a^2/4 $$ since $0\leq x\leq a$.

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