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I got an initial mean $\mu_1$ and std $\sigma_1$ by sampling samples, these samples are generated by an unknown distribution and later I drop these samples. Then I sampled some samples and got the mean $\mu_2$ and std $\sigma_2$ from the new sample and I kept the new samples. So how can I get the new std?

My idea is $(\sigma_1+\sigma_2)/2$ but I think this result is biased.

Another idea is $\mu_3=(\mu_1+\mu_2)/2$, and I reconstruct the previous samples by making an array with all items are $\mu_1$. Then combine the previous samples with the new samples and use the $\mu_3$ to get the $\sigma_3$.

Update: I clarify my question

I want to estimate mean and variance from unknow sample, due to memory cost, I can sample the data and get the mean and variance and then drop these samples, and then I sample data from the distribution and compute the mean and variance of the new samples, before dropping the new sample, I want to estimate the mean and variance of the distribution based on the current mean and variance and that of the previous step.

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  • $\begingroup$ Do you have sample size? $\Sigma$ is just std or variance-covariance matrix? $\endgroup$
    – user158565
    Commented Aug 8, 2019 at 4:00
  • $\begingroup$ @user158565 I have the sample size. $\Sigma$ is std. $\endgroup$
    – GoingMyWay
    Commented Aug 8, 2019 at 5:09
  • $\begingroup$ Perhaps I don't understand what you're trying to do, but why can't you just find $\bar{x}$ and $s^2$ for the second sample? $\endgroup$
    – Dave
    Commented Aug 8, 2019 at 14:42
  • $\begingroup$ Isn't the "new std" $\sigma_2$ by definition? If you dropped samples, surely that means you don't want to include them in your estimates, right? If all you want to do is combine the two sample sets, then your question is answered at stats.stackexchange.com/questions/51622, stats.stackexchange.com/questions/43159, stats.stackexchange.com/questions/30495, and other places. $\endgroup$
    – whuber ♦
    Commented Aug 8, 2019 at 15:07
  • $\begingroup$ @Dave I have to combine the first and the second round results. $\endgroup$
    – GoingMyWay
    Commented Aug 9, 2019 at 6:33

1 Answer 1

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Let $\bar X$ and $S$ be sample mean and standard deviation. (In statistics, $\mu$ and $\sigma$ are used for parameters).

Your problem can be resolved based on two formula:

$$\bar X = \frac {\sum_{i=1}^n X_i}n$$ $$S=\sqrt{\frac {\sum_{i=1}^nX_i^2-\frac {(\sum_{i=1}^nX_i)^2} n}{n-1}}$$

Applying them to two sets of $\bar X$ and $S$ from two samples, you can get two sets of $\sum_{i=1}^n X_i$ and $\sum_{i=1}^nX_i^2$. Adding them together you get new set $\sum_{i=1}^n X_i$ and $\sum_{i=1}^nX_i^2$ over the two samples. Applying two formula again, you get what you want.

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