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Can someone explain why: Given a set of n independent observations Z1...Zn, each with variance K. The variance of the mean is K/n?


marked as duplicate by gung, John, Christoph Hanck, Andy, Nick Cox Jan 17 '16 at 14:13

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Variance captures the spread of the data. I am not providing mathematical proof, but a pictorial solution for better understanding with a special case where we have only 2 independent variables X and Y, with variance K.

enter image description here

Without loss of generality, assume that K = 4 units. Here, the two bigger circles represent the spread of the data for random independent variables X and Y. Let P1, be a value generated by the random variable X. Then, for all the values generated by Y, when their mean is computed with that of P1, the result is another set of values represented by the smaller circle in the center. Clearly, the variance of the center sphere is K/2 = 2 units.

Hope this helps in understanding the concept.

  • 1
    $\begingroup$ Your picture appears to depict ranges of the variables, not their variances. What do ranges have to do with variances? $\endgroup$ – whuber Jan 17 '16 at 1:22

The mathematical proof for determining variance of the mean of independent random variables, with same variance is very simple:

Suppose we have $n$ independent random variables ($X_{1},X_{2},...,X_{n}$) with same variance $\sigma^{2}$ then,

$\bar{X} = \dfrac{X_{1} + X_{2} + ... + X_{n}}{n}$

$ \begin{align*} var(\bar{X}) &= var(\dfrac{X_{1} + X_{2} + ... + X_{n}}{n}) \\ &= \dfrac{1}{n^{2}} \times var(X_{1} + X_{2} + ... + X_{n}) \\ &= \dfrac{1}{n^{2}} \times (var(X_{1}) + var(X_{2}) + ... + var(X_{n})) \\ &= \dfrac{1}{n^{2}} \times n\sigma^{2} \\ &= \dfrac{\sigma^{2}}{n} \end{align*} $

Note that, in this derivation makes use of a couple of concepts :

  • A property of variance : $var(aX + b) = a^{2}var(X)$, where $a$ and $b$ are any constants.
  • A property of independent random variables : $var(\sum(X_{i})) = \sum(var(X_{i}))$

Thats it !!! You`re done.


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