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Let's say I have a random variable X, for which the following properties hold: $E[X]=2,\ D[X]=1.6$. Using Chebyshev's inequality, my task is to put an upper bound on $P(|X-5|\ge3)$. Using Chebyshev's inequality, I know that $P(|X-E[X]|\ge\epsilon)\le\frac{D[X]}{\epsilon^2}, \forall \epsilon > 0$. As you can see, in this problem $E[X]\ne5$, so I'm not sure what to do with it.

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  • $\begingroup$ Presumably $D[X]$ is supposed to be the variance of $X$ $\endgroup$ – Henry Jun 12 '17 at 6:59
  • $\begingroup$ Yes, indeed. E[X] is the average value and D[X] is the variance (it's also sometimes noted as $\sigma^2$. $\endgroup$ – SalysBruoga Jun 12 '17 at 7:07
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    $\begingroup$ The best upper bound you can get for $P[|X-5|\geq 3]$ is the trivial bound of $1$, which is achieved by a random variable $X$ that takes two values $\{8, x\}$ with probabilities $p, 1-p$, with parameters $x,p$ chosen to satisfy $E[X]=2$ and $Var(X)=1.6$. Likely this was just a typo and the problem meant to ask you for an upper bound on $P[|X-2|\geq \epsilon]$. [Aside: I have never seen $D[X]$ used for variance.] $\endgroup$ – Michael Jun 12 '17 at 7:40

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