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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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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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