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Science Forum Index » Statistics - Math Forum » Chebychev's inequality
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Message |
| Prat |
 Posted: Tue Jan 09, 2007 4:46 pm |
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Guest
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Hello,
I am a novice in Stat/Probability.
My question is:
Consider a random variable z following an arbitrary probability
distribution for which we know the mean mu and variance s^2
Quote: From Chebychex, we know that, for k >0,
P(z-mu >= t) <= s^2 / (s^2 + t^2)
This is equivalent to:
1 - P(z-mu < t) <= s^2 / (s^2 + t^2),
or
1 - s^2 / (s^2 + t^2) <= P(z-mu < t) ,
Is is then correct to say that?
1 - s^2 / (s^2 + t^2) <= P(z-mu > t)
Could you please give me a help on this?
Pratim |
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