The usual estimators of the mean and variance can be manipulated to get what you want. The estimator of the mean, for example, is simply the sum of all samples divided by the number of samples. Assuming that  you store the number of samples along with the estimated means and variances, the solution is simple: multiply the mean for each segment by the number of samples in each segment and add. Then divide by the total number of samples.

In numbers: segment A has 100 samples and mean 11, segment B has 120 samples and mean 10.  Then the sum of all samples in segment A is 100 x 11 and in segment B is 120 x 10. The value of the estimator for the two segments would be (100 x 10 + 120 x 11) divided by (100 + 120). In your case, it seems that all segments have the same length, hence it is even simpler.

The variance can be expressed as the mean of the squares minus the square of the means. The new quantity here is the mean of the squares, which can be handled as explained above (replacing samples by squared samples).

More interesting is the problem of combining the results of 2 (or more) measurements, assuming that they are unbiased but correlated. The best linear unbiased estimate (BLUE) provides a nice & simple solution to this problem. It can easily be expressed in terms of the covariance matrix and I'm sure you'll be able to easily find all the details about it (just lookup BLUE, "best linear unbiased estimator").

The usual estimators of the mean and variance can be manipulated to get what you want. The estimator of the mean, for example, is simply the sum of all samples divided by the number of samples. Assuming that  you store the number of samples along with the estimated means and variances, the solution is simple: multiply the mean for each segment by the number of samples in each segment and add. Then divide by the total number of samples.

In numbers: segment A has 100 samples and mean 11, segment B has 120 samples and mean 10.  Then the sum of all samples in segment A is 100 x 11 and in segment B is 120 x 10. The value of the estimator for the two segments would be (100 x 10 + 120 x 11) divided by (100 + 120). In your case, it seems that all segments have the same length, hence it is even simpler.

The variance can be expressed as the mean of the squares minus the square of the means. The new quantity here is the mean of the squares, which can be handled as explained above (replacing samples by squared samples).

More interesting is the problem of combining the results of 2 (or more) measurements, assuming that they are unbiased but correlated. The best linear unbiased estimate (BLUE) provides a nice & simple solution to this problem. It can easily be expressed in terms of the covariance matrix and I'm sure you'll be able to easily find all the details about it (just lookup BLUE, "best linear unbiased estimator").

As  Paulo Jorge S. G. Ferreira well said above, a weighted average is all you need. Best of luck.

Deal Paulo Jorge!

Thank you very much for your previous answer, which was very helpful!

In my case, for the mean value everything it works without objections. Unfortunately I could not get the standard deviation for all observations based on partial standard deviation calculated and recorded every 1 second (1000 points)

In Matlab, you can write it in this way:

s1 = [1 2 4 4 8];

s2 = [1 2 5 3 1];

m_total = mean ([s1 s2]);

m_parts = (mean (s1) + mean (s2)) / 2;

m_total = mean_parts

this two values are equal.

Could I ask you to write an equation / formula, based on which you can determine the standard deviation for the entire observation?

I remember that I have only data:

m_i = mean (s_i);

var_i = var (s_i);

std_i = std (s_i);

Thank you very much for your help!

The maths is very straightforward. I'm assuming that the random variables are drawn from the same population, i.e. distributed from the same pdf, and uncorrelated. Then concentrating on the 2nd moment only (since you can obtain the overall variance from this given the overall mean), which I label u2 (identically the mean of the square of your data) you have, by expanding the definition of u2 in terms of appropriate sums:

u2(([x y]) = (Nx-1)/(N-1)*u2(x) + (Ny-1)/(N-1)*u2(y)

where Nx is the number of x samples, Ny the number of y samples and N = Nx+Ny. u2(x) is the 2nd moment of x etc.

I assume an unbiased estimator here, hence the Nx-1, N-1 etc.

Then the standard deviation is the square root of (2nd moment - mean^2), where the overall mean, also called the 1st moment, is obtained by similar procedure:

u1([x y]) = Nx/N*u1(x) + Ny/N*u1(y)