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1. 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.