I think the question is valuable for the community as I could not find out any close related topic on stackexchange or even internet.
We have a sample of observations and we decided to replicate some of them nevertheless each of the replication have a weight and all replications weights of a specific observation have to sum to 1.
I need a formal proof that the mean and central moments are the same for the original sample and for the new replicated one when taking into account such weights.
I already have done some work, still I am looking for something more formal:
The main assumption for this example is that the weights for each replicated observation sum to one $\sum\limits_{i=1}^{n_1} w_i = 1, ..., \sum\limits_{i=1}^{n_m} w_i = 1$:
$$ \begin{aligned} \mu_{weighted} &= \frac{\overbrace{\sum\limits_{i=1}^{n_1} w_i x_1}^{x_1} + ... + \overbrace{\sum\limits_{i=1}^{n_m} w_i x_n}^{x_n}}{\underbrace{\sum\limits_{i=1}^{n_1} w_i}_{1} + ... + \underbrace{\sum\limits_{i=1}^{n_m} w_i}_{1}} = \mu = \frac {\sum_{i=1}^n {x_i}}{n} \end{aligned} $$
$$ \begin{aligned} \hat \mu_n^\mathrm{weighted} &= \frac{\overbrace{\sum\limits_{i=1}^{n_1} w_i \left(x_1 - \mu\right)^n}^{\left(x_1 - \mu\right)^n} + ... + \overbrace{\sum\limits_{i=1}^{n_m} w_i \left(x_n - \mu\right)^n}^{\left(x_n - \mu\right)^n}}{\underbrace{\sum_{i=1}^N w_i}_{1} + ... + \underbrace{\sum_{i=1}^N w_i}_{1}} = \hat \mu_n = \frac{\sum\limits_{i=1}^N \left(x_i - \mu\right)^n} N \end{aligned} $$
Thanks for a support or for references