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Actually I joined a few days ago and found a guy I really like. Met my girlfriend on Cam Voice, been together 4 months, going well so far!… continue reading »


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A pretty simple man with simple wants in life but not.. … continue reading »


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The definition for unbiased estimates of mean($\bar x$) and variance($\sigma^2$) for a sample of size n are: $$ \bar x_n=\frac1n\sum_^nx_k, $$ and $$ \bar\sigma^2_n=\frac1\sum_^n(x_k-\bar x_n)^2 $$ Define the recursion variables $$ M_n = n \bar x_n=\sum_^nx_k, $$ and $$ S_n = (n-1)\bar\sigma^2_n=\sum_^n(x_k-\bar x_n)^2 $$ The recursion relation for $M_$ is obvious $$ M_ = M_n x_ $$ and the recursion relation for $S_n$ is obtained via $$ S_ = (x_-\bar x_)^2 \sum_^n(x_k-\bar x_)^2\phantom\ \phantom = (x_-\bar x_)^2 \sum_^n(x_k-\bar x_n \bar x_n-\bar x_)^2\ \phantom = (x_-\bar x_)^2 \sum_^n(x_k-\bar x_n)^2 2(\bar x_n-\bar x_)\sum_^n(x_n-\bar x_n) \sum_^n(\bar x_n -\bar x_)^2\ $$ And since $$S_n = \sum_^n(x_k-\bar x_n)^2$$ $$\sum_^n(\bar x_n-\bar x_)^2 = n(\bar x_n-\bar x_)^2$$ and $$\sum_^n(x_k-\bar x_n) = 0$$ this simplifies to $$ S_ = S_n (x_-\bar x_)^2 n(\bar x_n -\bar x_)^2 $$ Now, this recursion relation has the nice property that it $S_n$ a sum of squared terms, and thus cannot be negative.… continue reading »


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