I sample $n$ points from standard normal and need the mean and variance of 3rd and 4th sample cumulants.
@JimB suggested that variance of 4th sample cumulant is given by the expression below. This seems like a very algebraically nice expression, is there way to derive/verify it with the help of Mathematica?
$$ \frac {24(n-1)((n-6)n+24)}{n^4} $$
Edit, Sep 5
The following gives a way to verify the formula for a fixed $n$, but not helpful for deriving the formula in terms of $n$
n = 5;
Clear[v];
rvec = Array[v, n];
sampleDist = ProductDistribution @@ Table[NormalDistribution[], n];
cumulantDist =
TransformedDistribution[Cumulant[rvec, 4],
rvec \[Distributed] sampleDist];
Variance[cumulantDist] == (24 (n - 1) ((n - 6) n + 24))/n^4