# What is the reason to define the variance? [duplicate]

Possible Duplicate:
Usefulness of Variance

i know Standard deviation is equal to a constant times the shortest distance from the point $(x_1,x,_2,\ldots,x_n)$ to a line $(r_1,r_2,r_3,\ldots,r_n)$ where the r is equal provide that these objects is in n-dimensional euclidean metric space, it is technically called the geometrical interpretation of the standard deviaiton.

My question is, i know the standared deviation have its remarkable significance in both probability and statistics, so what is the reason to define another term called the variance

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## marked as duplicate by Ｊ. Ｍ., leonbloy, William, Rahul, t.b.Sep 11 '12 at 16:16

From where did you get what you know about "..standard deviaton is equal to a constant times the shortest distance from $\ldots$"? –  Dilip Sarwate Sep 4 '12 at 1:19

$\newcommand{\var}{\operatorname{var}}$ Variances are additive: If $X$ and $Y$ are independent random variables, then $$\var(X+Y) = \var(X)+\var(Y),\tag{1}$$ and similarly for more than two random variables.
One of the most obvious measures of dispersion is the average distance between $x_i$ and the sample mean $(x_1+\cdots+x_n)/n$. But nothing closely connected to it has this nice additive property. Sample 1000 members of a population, chosen independently of each other. Measure their height or their income or their cholesterol level or their proximity to Mount Rushmore, and add them up. What's the variance of the sum? Just the sum of the variances! Without this easy identity, one could never apply the central limit theorem. When you toss a coin 1800 times, what's the probability that the number of heads you get is between 1785 and 1831? Just use the central limit theorem! In the case of coin tosses, this goes back to Abraham de Moivre in the early 18th century. In the case of more complicated distributions, I think it's around 1930 or so. It would be impossible without the identity $(1)$ above.
Variance is structurally much better behaved than standard deviation. For example, $$\frac{1}{n-1}\sum_1^n (x_i-\overline{x})^2$$ is a pleasant unbiased estimator of the variance. Unbiased estimators of standard deviation are dependent on distribution, and can be very complicated.
OK, using $\frac{1}{n}$ is arguably "better." A minimum variance estimator for standard deviation brings up similar issues. –  André Nicolas Sep 4 '12 at 2:31