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Forums
Mathematics
Set Theory, Logic, Probability, Statistics
Weak Convergence to Normal Distribution
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[QUOTE="Opalg, post: 6777771, member: 703375"] I am not a probabilist, but I think that your formula should read $\frac{1}{\sqrt{n}} \Bigl(\left(\sum_{m=1}^{n} X_m\right)- \frac{n}{2}\Bigr)$. If $m$ is large, then $X_m$ takes each of the values $0$ and $1$ with a probability close to $\frac12$ (and the large value $m$ with a very small probability $m^{-2}$). So the mean value of $X_m$ will be close to $\frac12$, and the mean value of $S_n = \sum_{m=1}^{n} X_m$ will be close to $\frac n2$. This seems to make it plausible that your formula should somehow relate to the central limit theorem. [/QUOTE]
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Weak Convergence to Normal Distribution
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