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I recently got a tute question which I don't know how to proceed with and I believe that the tutor won't provide solution... The question is

Pick a real number randomly (according to the uniform measure) in the interval $[0, 2]$. Do this one million times and let $S$ be the sum of all the numbers. What, approximately, is the probability that
a) $S\ge1,$
b) $S\ge0.001,$
c) $S\ge0$?
Express as a definite integral of the function $e^\frac{-x^2}{2}$.

Can anyone show me how to do it? It is in fact from a Fourier analysis course but I guess I need some basic result from statistcs which I am not familiar with at all..

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    @marty: not much: (a) would become $\frac12$, (c) would still be $1$ and (b) would still be very close to $1$2012-08-19

2 Answers 2

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You have $1\, 000\, 000$ independent random variables $X_j$ with uniform distribution on the interval $[0,2]$. Such a random variable has mean $\mu = 1$ and variance $\sigma^2 = 1/3$. The central limit theorem says that if $S_N$ is the sum of $N$ independent random variables with the same distribution, having mean $\mu$ and variance $\sigma^2$, then in the limit as $N \to \infty$, the distribution of $(S_N - N \mu)/(\sqrt{N} \sigma)$ approaches the standard normal distribution. Thus, for any $z$,

$\text{Prob} \left( \frac{S_N - N\mu}{\sqrt{N} \sigma} \ge z \right) \to \int_z^\infty \frac{e^{-t^2/2}}{\sqrt{2\pi}} dt$

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    I don't think changing the question to the average would make the central limit theorem any more useful: the approximation might be correct for (a) but the exact answer is in any case obvious by symmetry, while (b) and (c) are still extreme.2012-08-20
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Let's call $S_n$ the sum of the first $n$ terms. Then for $0 \le x \le 1$ it can be shown by induction that $\Pr(S_n \le x) = \dfrac{x^n}{2^n \; n!}$

So the exact answers are

a) $1 - \dfrac{1}{2^{1000000} \times 1000000!}$

b) $1 - \dfrac{1}{2000^{1000000} \times 1000000!}$

c) $1$

The first two are extremely close to 1; the third is 1. The central limit theorem will not produce helpful approximations here, so you may have misquoted the question.

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    Yes, you are absolutely right, I was thinking of the full range, which is irrelevant.2012-08-19