Kurtosis and Skewness for 2 or more samples' combination

Question

Question:

Can we calculate kurtosis and skewness of 2 or more combined samples, given each samples' mean, std, sample size and kurtosis/skewness?

Let's say we have sample size, sample mean, sample standard deviation and sample kurtosis of 2 samples($\{x_1,x_2,...,x_n\}$ and $\{y_1,y_2,...,y_m\}$) , can we calculate the kurtosis of the combination of these 2 samples $\{x_1, x_2,...x_n,y_1, ...,y_m\}$?

The problematic thing here is we focusing on calculating skewness/kurtosis by subsamples' statistics instead of original sample/subsample. In other word, we need to get the skewness/kurtosis without any touch of original data! This is reasonable in real industrial practice due to limits on RAM and CPU times.

Further Question:

I have proved that:

1. For 2 independent samples, if we have sample size, sample mean, sample deviation of each sample, we can calculate the mean and std of combined sample.
2. For 2 independent samples($\mathrm{pvctr_1}= \frac{\# clicks}{\# expos}=\frac{\sum_{i=1}^{n} x_{1i}}{\sum_{i=1}^{n} y_{1i}}, \mathrm{pvctr_2}= \frac{\# clicks}{\# expos}=\frac{\sum_{j=1}^{m} x_{2j}}{\sum_{j=1}^{m} y_{2j}}$), if we have sample size($n$ for sample 1,$m$ for sample 2), numerator/denominator's sample mean/sample deviation for each sample, we can not calculate the mean and std of combined sample's $\mathrm{pvctr}$. If we also know the covariance between numerator and denominator in each sample, then we can.

I was wondering if there exists any theorem gives the sufficient sample statistics of each sample, to help us calculating sample statistics for a combined sample?

With all due respect, I think I should emphasize the last question of the post:

I was wondering if there exists any theorem gives the sufficient sample statistics of each sample, to help us calculating sample statistics for a combined sample?

I am asking a most powerful solution to this kind of question, instead a general way of thinking. For me, I did know how to calculate mean/std/skewness/kurtosis based on subsamples' up to 4th raw/central moments. But, there exists duplicate/useless information in up to 4th raw/central moments, which means we don't need all of them to calculate combined samples' statistics. Thus, I want to know the "Sufficient Statistics of subsamples" for calculating combined sample particular statistics, and then I can keep my solution extremely tiny and powerful.

Answer 1

As a general rule, you can compute the $k$th order sample moments of a pooled sample so long as you have all the sample moments of the underlying subgroups up to order $k$. So, in order to compute the sample kurtosis of the pooled sample you need to have the sample mean, sample variance, sample skewness and sample kurtosis of the two subgroups that are to be pooled. (In your question you missed the skewness, so technically, the answer to your question is no; you can't get the pooled sample kurtosis with just the statistics you mention.)

Fortunately, you needn't re-invent the wheel here, since a brilliant and devoted statistician has already done this work for you in the utilities package in R. Statistical problems of this kind are automated in the sample.decomp function in that package. This function can compute pooled sample moments from subgroup moments, or compute missing subgroup moments from the other subgroup moments and pooled moments. It works for any decompositions up to fourth order ---i.e., decompositions of sample size, sample mean, sample variance/standard deviation, sample skewness, and sample kurtosis.

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How to use the function: Here we give an example where we use the function to compute the sample moments of the pooled data from two subgroups. First we randomly generate some data from a fixed distribution and find the sample moments of the two subgroups.

#Generate some example data
set.seed(1)
DATA.A <- rgamma(n = 12, shape = 2, scale = 6)
DATA.B <- rgamma(n = 20, shape = 2, scale = 6)

#Compute and store the sample moments
library(utilities)
MOMENTS.A <- moments(DATA.A, include.sd = TRUE)
MOMENTS.B <- moments(DATA.B, include.sd = TRUE)

#Show the moments of dataset A
MOMENTS.A
        n sample.mean sample.sd sample.var sample.skew sample.kurt NAs
DATA.A 12    11.94847  6.946869   48.25898   0.4535123    1.738779   0

#Show the moments of dataset B
MOMENTS.B
        n sample.mean sample.sd sample.var sample.skew sample.kurt NAs
DATA.B 20    11.12291  6.301247   39.70571   0.2532274    1.731373   0

Now that we have two subgroups with known sample moments (up to fourth order), let's find the moments of the pooled sample using the sample.decomp function.

#Compute sample decomposition
N    <- c(MOMENTS.A$n, MOMENTS.B$n)
MEAN <- c(MOMENTS.A$sample.mean, MOMENTS.B$sample.mean)
VAR  <- c(MOMENTS.A$sample.var,  MOMENTS.B$sample.var)
SKEW <- c(MOMENTS.A$sample.skew, MOMENTS.B$sample.skew)
KURT <- c(MOMENTS.A$sample.kurt, MOMENTS.B$sample.kurt)
sample.decomp(n = N, sample.mean = MEAN, sample.var = VAR,
              sample.skew = SKEW, sample.kurt = KURT, 
              names = c('DATA.A', 'DATA.B'), include.sd = TRUE)

            n sample.mean sample.sd sample.var sample.skew sample.kurt
DATA.A     12    11.94847  6.946869   48.25898   0.4535123    1.738779
DATA.B     20    11.12291  6.301247   39.70571   0.2532274    1.731373
--pooled-- 32    11.43250  6.451729   41.62481   0.3535067    1.791869

We can confirm that these moments match the results from first combining the subgroups and then computing the moments of the pooled data directly.

#Combine the two subgroups
POOLED <- c(DATA.A, DATA.B)

#Compute the moments of the pooled sample
MOMENTS.POOLED <- moments(POOLED, include.sd = TRUE)

#Show the pooled moments
MOMENTS.POOLED
        n sample.mean sample.sd sample.var sample.skew sample.kurt NAs
POOLED 32     11.4325  6.451729   41.62481   0.3535067    1.791869   0