# Chebyshev's inequality

• Jun 27th 2013, 12:30 AM
downthesun01
Chebyshev's inequality
Use Chebyshev's inequality to get an upper bound on the probability $P\Big[|X-E[X]|>2\sqrt{Var[X]}\Big]$, and also calculate the exact probability.

X is the result of a fair dice toss, 1,2,3,4,5,6, each with a probability of $\frac{1}{6}$

I'm not sure what it means by "get an upper bound on the probability."

This is what I did, and it matches the manual's solution of what the "exact probability" is:

$E[X]=\frac{1+2+3+4+5+6}{6}=\frac{7}{2}$

$E[X^2]=\frac{1+4+9+16+25+36}{6}=\frac{91}{6}$

$Var[X]=E[X^2]-E[X]^2=\frac{91}{6}-\frac{49}{4}=\frac{35}{12}$

$P\left[|X-\frac{7}{2}|>2\sqrt{\frac{35}{12}}\right]=$ $1-P[0.0843\leq X \leq 6.916]=1-P[1,2,3,4,5,6]=1-1=0$

$0\leq \frac{1}{r^2}=0\leq\frac{1}{4}$
• Jun 27th 2013, 03:14 AM
chiro
Re: Chebyshev's inequality
Hey downthesun01.

I think what its asking you to do is use the inequality (where k = 2) and to compare with (and state) the exact probability that you have figured out.

The bound is 1/k^2 which is 1/4 which represents a theoretical upper bound given Chebyshevs Inequality however the true value (i.e. exact) is a lot smaller (i.e. 0) and still remains in the bound (since 0 < 1/4).
• Jun 27th 2013, 03:19 PM
downthesun01
Re: Chebyshev's inequality
Okay, that makes sense now. Thanks