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Using Chebychev's inequality, Show that if $$P(a\le X \le b)=1, \infty<a<b<\infty$$ then $$var(x)\le \frac{(b-a)^2}{4}$$ Answer:- I tried to do in this way,

Let us assume $X$ to have a uniform distribution over the interval [a,b] with the p.d.f $$f(x)=\frac{1}{b-a}, a\le X \le b$$ It is now quite obvious that $P(a\le X \le b)=1$ We have $$Mean=\mu=\frac{b+a}{2}$$ Now, $$P(a\le X \le b)=P(a-\frac{b+a}{2} \le X-\mu \le b-\frac{b+a}{2})=P(\frac{a-b}{2}\le X-\mu \le \frac{b-a}{2})=P(|X-\mu|\le\frac{b-a}{2})=1 ..........(*)$$ Again by Chebychev's inequality, $$P(|X-\mu|\le k)\ge 1-\frac {Var(X)}{k^2}$$ So, $$P(|X-\mu|\le\frac{b-a}{2} )\ge 1-\frac {Var(X)}{(\frac{b-a}{2})^2}......(**)$$ From $(*)$ and $(**)$ we have $$\frac {Var(X)}{(\frac{b-a}{2})^2}\le1\Rightarrow Var(X)\le(\frac{b-a}{2})^2$$ Hence the result.

Am I correct in my approach ? I will be happy to be pointed out in case of any mistake.

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    $\begingroup$ You must do it for general $X$, right? Why you assume $X$ to be uniformly distributed? You just showed it for one specific case, didn't you? $\endgroup$ – Shashi Feb 1 '18 at 0:58
  • $\begingroup$ Haven't you just concluded that the variance is non negative? $\endgroup$ – user251257 Feb 1 '18 at 1:07
  • $\begingroup$ Yes you are absolutely correct. I think there will be some another way to show it for a general $X$ satisfying the above criteria. However, I am getting it. $\endgroup$ – user440191 Feb 1 '18 at 1:09
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I am not sure why you need Chebychev.

Notice that since $X$ takes values in an compact interval $[a,b]$, it has a mean, say $\mu$, and variance, say $\sigma^2$. Let $c=(a+b)/2$. Then, we have $|X-c| \le (b-a)/2$.

By the minimizing property of the mean, we have

$$ \sigma^2 = E[ (X-\mu)^2 ] \le E [ (X - c)^2 ] \le \frac{(b-a)^2}{4}.$$

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  • $\begingroup$ You are absolutely right. $\endgroup$ – user440191 Feb 1 '18 at 1:54

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