Recursive formula for variance

Recall that, for every $n\geqslant1$, $$ \bar x_n=\frac1n\sum_{k=1}^nx_k, $$ and $$ \bar\sigma^2_n=\frac1n\sum_{k=1}^n(x_k-\bar x_n)^2=\frac1n\sum_{k=1}^nx_k^2-(\bar x_n)^2. $$ Hence simple algebraic manipulations starting from the identities $$ (n+1)\bar x_{n+1}=n\bar x_n+x_{n+1}, $$ and $$ (n+1)(\bar\sigma^2_{n+1}+(\bar x_{n+1})^2)=n(\bar\sigma^2_n+(\bar x_n)^2)+x_{n+1}^2, $$ lead to $$ \bar x_{n+1}=\bar x_n+\frac{x_{n+1}-\bar x_n}{n+1}, $$ and $$ \bar\sigma^2_{n+1}=\bar\sigma^2_n+(\bar x_n)^2-(\bar x_{n+1})^2+\frac{x_{n+1}^2-\bar\sigma^2_n-(\bar x_n)^2}{n+1}. $$ Thus, $(n,\bar x_n,x_{n+1})$ yield $\bar x_{n+1}$ and $(n,\bar\sigma^2_n,\bar x_n,\bar x_{n+1},x_{n+1})$ yield $\bar\sigma^2_{n+1}$.

There are two problems in the preceding answer, the first being the formula for the variance is incorrect(see the formula below for the correct version) and the second is that the formula for the recursion ends up subtracting large, nearly equal, numbers.

The definition for unbiased estimates of mean($\bar x$) and variance($\sigma^2$) for a sample of size n are: $$ \bar x_n=\frac1n\sum_{k=1}^nx_k, $$ and $$ \bar\sigma^2_n=\frac1{n-1}\sum_{k=1}^n(x_k-\bar x_n)^2 $$

Define the recursion variables

$$ M_n = n \bar x_n=\sum_{k=1}^nx_k, $$ and $$ S_n = (n-1)\bar\sigma^2_n=\sum_{k=1}^n(x_k-\bar x_n)^2 $$

The recursion relation for $M_{n+1}$ is obvious $$ M_{n+1} = M_n + x_{n+1} $$ and the recursion relation for $S_n$ is obtained via $$ S_{n+1} = (x_{n+1}-\bar x_{n+1})^2+\sum_{k=1}^n(x_k-\bar x_{n+1})^2\phantom{XXXXXX}\\ \phantom{S_{n+1}} = (x_{n+1}-\bar x_{n+1})^2+\sum_{k=1}^n(x_k-\bar x_n+\bar x_n-\bar x_{n+1})^2\\ \phantom{S_{n+1}XXXXXXXXXX} = (x_{n+1}-\bar x_{n+1})^2+\sum_{k=1}^n(x_k-\bar x_n)^2+2(\bar x_n-\bar x_{n+1})\sum_{k=1}^n(x_n-\bar x_n) + \sum_{k=1}^n(\bar x_n -\bar x_{n+1})^2\\ $$ And since $$S_n = \sum_{k=1}^n(x_k-\bar x_n)^2$$ $$\sum_{k=1}^n(\bar x_n-\bar x_{n+1})^2 = n(\bar x_n-\bar x_{n+1})^2$$ and $$\sum_{k=1}^n(x_k-\bar x_n) = 0$$ this simplifies to $$ S_{n+1} = S_n+(x_{n+1}-\bar x_{n+1})^2 +n(\bar x_n -\bar x_{n+1})^2 $$

Now, this recursion relation has the nice property that it $S_n$ a sum of squared terms, and thus cannot be negative. Written in terms of $M_n$ and $S_n$, the recursion relations are: $$ M_{n+1} = M_n + x_{n+1} $$ $$ S_{n+1} = S_n+\left(x_{n+1}-\frac{M_n+x_{n+1}}{n+1}\right)^2 +n\left(\frac{M_n}{n} -\frac{M_n+x_{n+1}}{n+1}\right)^2 $$ and we can further simplify the recursion relation for $S_n$ to \begin{eqnarray} S_{n+1} &= S_n + \left(\frac{n x_{n+1}-M_n}{n+1}\right)^2+n\left(\frac{M_n-n x_{n+1}}{n(n+1)}\right)^2\\ &=S_n+ \left(1+\frac1n\right)\left(\frac{n x_{n+1}-M_n}{n+1}\right)^2\\ &=S_n+ \frac{(n x_{n+1}-M_n)^2}{n(n+1)} \end{eqnarray}

So we have the simple recursion relations: $$ M_{n+1} = M_n + x_{n+1} $$ $$ S_{n+1} = S_n + \frac{(n x_{n+1}-M_n)^2}{n(n+1)}$$

with the mean given by $$\bar x_n = \frac1n M_n$$ and the unbiased estimate of the variance is given by $$\sigma_n^2 = \frac1{n+1}S_n$$.