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A simple proof of Gauss's inequality
Published online by Cambridge University Press: 11 August 2014
Extract
Consider a continuous probability distribution of a variable x measured from its mean in units of its standard deviation, and suppose that deviations from the mean are taken irrespective of sign; let px represent the ordinate applicable to x (this ordinate is, of course, the sum of the original ordinates at + x and — x). We then have
, and we write .
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- Copyright © Institute of Actuaries Students' Society 1947
References
page 40 note * This is Winkler's extension of Gauss's inequality [Ed.].
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