left-hand-tailed chebyshev (Cantelli's) inequality

kpjoyce

New Member
I'd like to use the one-tailed chebyshev inequality to help me describe some weather data. The proper inequality as I understand it is:
P{X-E(X)>=k(sigma)}<=1/(1+k^2)
Which describes the right-hand tail of the distribution (probability that values are above a bound).
Can someone help me prove to myself that a similar (or the same) equality exists for the left-hand tail (probability that values are below a lower bound)?
P{E(X)-X>=k(sigma)}<=???

Thanks-
Kevin

edit ... well, this is not homework. Still, any help?

Xenu

New Member
It works the same for a lower bound, which can you can see if you choose k to be a negative value.

kpjoyce

New Member
I should have mentioned that the definition is only for k>0.

What you're saying would be true if we were talking about |X-E(X)|.

Thanks for thinking through it with me.

Xenu

New Member
deleted. Misunderstood the problem.

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