# Mean and variance of a normally distributed random number created from the average of a set of uniformly distributed random numbers

An old-fashioned way of generating normally distributed random numbers entailed setting each normally distributed random number equal to the average of a set of uniformly distributed random numbers, rescaled and shifted by a constant. Let x be a uniformly distributed random variable, i.e. p(x) = 0 if x < 0 and x ≥1; p(x) = 1 if 0 ≤ x <1.

How would I compute the mean and variance of the variable x?

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So if a set of uniform variables is a to b, a set of uniform variables from a+x to b-x has the same mean and variance as the set of uniform variables from a to b? – strimp099 Sep 20 '11 at 18:37
I don't follow what you mean by "a set of uniform variables is a to b" nor what distinction you are making between a mean or variance of a random variable and the mean or variance of a "set of [random] variables." Do you perhaps intend a set of observations of a single random variable? Another possible confusion is the use of "x" as a random variable in the question and--apparently--as a parameter in the comment. Is there any connection between the two? – whuber Sep 20 '11 at 19:33

If $X \sim {\rm Uniform}(0,1)$, then

$E[X] = \int_0^1 x \, dx = \frac12 x^2 |_0^1 = \frac12$

$E[X^2] = \int_0^1 x^2 \, dx = \frac13 x^3 |_0^1 = \frac13$

$V[X] = E[X^2] - \{ E[X] \}^2 = \frac13 - \frac14 = \frac1{12}$

I hope I don't have to demonstrate $E[aX+b]$ and $V[aX+b]$ for you.

FYI, there are much better ways to generate normal variates that do not require inversion of the normal CDF, such as the polar method.

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