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I was going through the proof of the Chebyshev Inequality here .

And I seem to be facing some trouble in the approximation stage. I can't seem to follow how $\epsilon$ has been approximated to $(t-\mu)$.

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4 Answers 4

up vote 7 down vote accepted

It is an inequality. The text in that document breaks up the flow slightly.

It should read like this:

$\int_{-\infty}^{\mu-\epsilon}(t-\mu)^2f_X(t)dt+\int_{\mu+\epsilon}^{\infty}(t-\mu)^2f_X(t)dt \ge \int_{-\infty}^{\mu-\epsilon}\epsilon^2f_X(t)dt+\int_{\mu+\epsilon}^{\infty}\epsilon^2f_X(t)dt$.

They did not replace $(t-\mu)^2$ with $\epsilon^2$. Rather, they exploited the inequality $(t-\mu)^2 \ge \epsilon^2$ to achieve the inequality I typed above.

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Four of us gave you the same correct explanation within a few minutes of each other so we all worked on this at the same time. So @Inquest why did you choose to give the check mark to Ed? Technically Peter was first, I was second and Ed was last. Is it that you found his explanation a little clearer? –  Michael Chernick Jul 17 '12 at 21:07
@MichaelChernick, For one, it was a correct explanation. Secondly, I appreciated the fact that it was no terse and I was pleased to find an explanation which didn't make me feel dumb and clarified that there was a break in the flow of the document. (Also, for some reason, I saw this one first). –  Inquest Jul 17 '12 at 21:15
I think there are some timestamp discrepancies. On my end, it shows my response at 32 minutes ago, @MichaelChernick's at 30 minutes ago, Peter's at 25 minutes ago and oen's at 31 minutes ago. I have also voted up all other answers, since they give the same explanation in slightly different ways, using different notation, which I think has educational value for someone learning the material. –  Arkamis Jul 17 '12 at 21:19

Notice that for $t\in (-\infty,\mu-\epsilon]\cup[\mu+\epsilon,\infty)$ that $(t-\mu)^2 \ge \epsilon^2$.

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Over the two semi infinite intervals of integration we have 1) in the first region t<μ-ϵ and 2)in the second region t>μ+ϵ. Both regions were cleverly chosen so the ϵ$^2$<(t-μ)$^2$. So the inequality is maintained with ϵ$^2$ replacing (t-μ)$^2$ and the rest should be easy for you.

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@Inquest. I think to be fair you need to give each of us an upvote. The checkmark is of course at your discretion. –  Michael Chernick Jul 17 '12 at 21:12
Although I am not bound by your opinions and see no reason why your solicitation should be fair, I did upvote all answers (including yours) when I came to understand them. –  Inquest Jul 17 '12 at 21:25
@Inquest Sure i understand and i appreciate that and what the ther respondents are doing. I am going to make sure that I upvote all the others as well. I think that is fair. We all came upon our answers independently. –  Michael Chernick Jul 17 '12 at 22:37

$\epsilon$ is not being approximated by $t - \mu$. What is happening is that for $t$ outside the interval $(\mu - \epsilon, \mu + \epsilon)$, $\epsilon^2 \leq (t - \mu)^2$ and hence the two integrals can be underestimated.

The reasoning "since $t \leq \mu - \epsilon \Rightarrow \epsilon \leq | t - \mu | \Rightarrow \epsilon^2 \leq (t - \mu)^2$" on the page you refer to, applies to the rewriting of the left integral, c.q., $t$ on the left of the interval $(\mu - \epsilon, \mu + \epsilon)$. A similar reasoning applies to the right integral, c.q., the right of that interval.

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