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If I change the power of the variance to $4$ or $6$, do the properties of the new formula remain the same as the old variance formula? If not what is the difference?

Variance formula:

$$\sigma^2=\frac{\sum(X-\mu)^2}{N}$$

Librecoin
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Victor
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  • @AndréNicolas - Can you list all the property that will be difference as a result in your answer? Thank you in advance! – Victor Apr 27 '13 at 21:14
  • @AndréNicolas - By the way, what is "variance of an independent sum is the sum of the variances." that you are talking about? – Victor Apr 27 '13 at 21:17

1 Answers1

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The algebra would become much less nice. We would lose the very useful fact that the variance of a sum of independent random variables is the sum of the variances. With the usual definition of variance, if $X$ and $Y$ are independent random variables, then $$\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y).$$ Sums, and more generally linear combinations of independent random variables are used a great deal in probability theory,

And with the altered definition, variance would become less useful as a measure of variability. For if we use the power $4$ or $6$, there will be an excessive sensitivity to infrequent large deviations from the mean.

André Nicolas
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  • what is" variance of an independent sum is the sum of the variances. – Victor Apr 27 '13 at 21:19
  • Can you show me the property of variance that "variance of an independent sum is the sum of the variances" in formula form please? – Victor Apr 27 '13 at 21:22