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A simple proof of Gauss's inequality

Published online by Cambridge University Press:  11 August 2014

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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 .

Type
Research Article
Copyright
Copyright © Institute of Actuaries Students' Society 1947

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References

page 40 note * This is Winkler's extension of Gauss's inequality [Ed.].