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Why is variance of a random variable bounded by $Var(X) \leq \mathbb{E}\left[\left(X-a\right)^2\right] $for any constant a ?

user100423
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1 Answers1

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Hint: Expand the right hand side and try to minimize the function with respect to $a$. Then check the sign of the $2$nd derivative to make sure you've minimized the function accurately.

Adding on to it: Write out the following and see how the sides of the expression compare. $$\mathbb{E}[(X-a)^2]=\mathbb{E}[(X-\mathbb{E}[X]+\mathbb{E}[X]-a)^2]=?$$In the above expression we've just added and subtracted $\mathbb{E}[X]$.

Sudarsan
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  • Can you elaborate more please?, i am not getting the idea – user100423 Dec 10 '13 at 02:33
  • You need to prove that the value that minimizes the expression $\mathbb{E}[(X-a)^2]$ is the expected value of $X$ thereby answering your question. – Sudarsan Dec 10 '13 at 02:35
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    Minimizing $E[(X-a)^2]$ is the standard way of approaching the problem, but if you had expanded out your displayed equation to get $$E[(X-a)^2] = E[(X-\mu)^2] + (\mu-a)^2 = \operatorname{var}(X) + (\mu-a)^2 \geq \operatorname{var}(X),$$ the 2nd derivative test etc could all have been avoided. – Dilip Sarwate Dec 10 '13 at 02:55