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Chebyshev’s inequality
Posted: 03 July 2009 03:02 AM   Ignore ]  
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Dear David,
I have the following question concerning the Chebyshev’s inequality:
Chebyshev’s inequality states that t e probability that X differs from its mean by at least k standard deviations is less than or equal to 1/k^2
It follows that the probability that X differs from its mean by less than k standard deviations is at least 1-1/k^2.
That is,
P( | X−μ |<kσ )≥1-1/k^2. DIV>
But in your Quant. notes it is stated that
P( | X−μ |<kσ )≥1/k^2. DIV>
Which formula is correct ?
Sincerely,
Alex.

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Posted: 03 July 2009 08:50 AM   Ignore ]   [ # 1 ]  
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Hi Alex,

The notes contained an error (which was fixed). Apologies. It is just as you say:

P( | X−μ |<kσ )≥1-1/k^2

(i.e., Gujarati 3.2 is also incorrect to gove P( | X−μ |<kσ ) <= 1-1/k^2)

Thanks, David

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Posted: 03 July 2009 09:01 AM   Ignore ]   [ # 2 ]  
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Thank you, David !

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