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In a setting where I have n clients each owns a subset of the dataset, each computes the mean and variance on his data and send them to the server. The server would average all means and variances. For mean, I simply do the averaging by: $\mu = \frac{\mu_1+\mu_2+.....+\mu_n}{n}$

However, for the variance, I can't figure out to average n variances. For only two I found this answer here

$\sigma^2 = \frac{1}{2}(\sigma^2_1 + \sigma^2_2) + (\frac{\mu_1-\mu_2}{2})^2$

But how about n variances? thank you

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Let's assume you have $n$ datasets, $x_k$ is the $k$th dataset. It's length we denote by $l_k$, its elements are $$x_k =\big\{x_k(1), x_k(2), \dots, x_k(l_k)\big\},$$

and its mean and variance are respectively \begin{align} \mu_k &= \frac{1}{l_k}\sum_{i=1}^{l_k} x_k(i),\\ \sigma_k^2 &= \frac{1}{l_k}\sum_{i=1}^{l_k} x_k(i)^2 - \mu_k^2. \end{align}

Then, noting that $l_k\mu_k = \sum_{i=1}^{l_k} x_k(i)$:

\begin{align} \mu &= \frac{1}{\sum_{k=1}^nl_k}\sum_{k=1}^n\sum_{i=1}^{l_k} x_k(i)\\ \mu &= \frac{1}{\sum_{k=1}^nl_k}\sum_{k=1}^nl_k\mu_k \end{align}

and in similar fashion

\begin{align} \sigma^2 &= \frac{1}{\sum_{k=1}^nl_k}\sum_{k=1}^n\sum_{i=1}^{l_k} x_k(i)^2 - \mu^2\\ \sigma^2 &= \frac{1}{\sum_{k=1}^nl_k}\sum_{k=1}^nl_k\left(\sigma^2_k + \mu_k^2\right) - \mu^2 \end{align}

When $l_k = N$ for all $k$ the above simplifies:

\begin{align} \mu &= \frac{1}{n}\sum_{k=1}^n\mu_k\\ \sigma^2 &= \frac{1}{n}\sum_{k=1}^n\left(\sigma^2_k + \mu_k^2\right) - \mu^2 \end{align}

however when the datasets have different sizes the above does not simplify, and you need to take that into account.

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