Minimum Numbers of Observations for Standardized Moment Calculations

Question

You can take the mean of any number of values, including just one value - in that case, the mean will just be equal to that value. Standardized means (standardized first moments) are always equal to zero.

You can't calculate variance (the second standardized moment) for only one value, though - you need a minimum of two values to calculate this moment. Since the variance of one number is zero, the standardized variance would be undefined since you'd have to divide by zero in that calculation.

I'm wondering if this pattern holds for higher-order moments. In other words, would it not make sense to calculate kurtosis (the 4^th moment) for three values? I know it's possible to calculate kurtosis for three values, but I'm not sold that doing so will actually tell you anything useful; also, this could be a degrees-of-freedom thing - perhaps there just aren't the degrees of freedom to calculate kurtosis for three values.

Is it reasonable to claim that, to calculate the n^th moment, you need a minimum of n values?

Furthermore, it's interesting that skewness is always 0 for groups of 2 observations (it's obvious why) and kurtosis is always 2 for groups of 3 observations (it's not obvious to me why). Higher-order n^th moments do not follow this pattern of always being the same when you have n - 1 observations, and here's some R code to prove it.

# Calculating Standardized Second Moments (Variances) for Different Groups of
# One Observation

One_Observation_Groups <- list(1, -100, 5, 25, 0)
sapply(One_Observation_Groups, function (x) {
  y <- sum(((x - mean(x)) / sd(x)) ^ 2)
})
# [1] NA NA NA NA NA

# Calculating Standardized Third Moments (Skewnesses) for Different Groups of
# Two Observations

Two_Observation_Groups <- list(c(1, 2), c(-3, 5), c(5, 500), c(1000, 1000000), c(-10, 0))
sapply(Two_Observation_Groups, function (x) {
  y <- sum(((x - mean(x)) / sd(x)) ^ 3)
})
# [1] 0 0 0 0 0

# Calculating Standardized Fourth Moments (Kurtoses) for Different Groups of
# Three Observations

Three_Observation_Groups <- list(c(1, 2, 3), c(4, 4, 9), c(10, 100, 1000), c(-1, 1, 20), c(-1000, 0, 25))
sapply(Three_Observation_Groups, function (x) {
  y <- sum(((x - mean(x)) / sd(x)) ^ 4)
})
# [1] 2 2 2 2 2

# Calculating Standardized Fifth Moments for Different Groups of Four
# Observations

Four_Observation_Groups <- list(c(1, 2, 3, 4), c(4, 4, 9, 4), c(10, 100, 1000, 10000), c(-1, 1, 20, -200), c(-1000, 0, 25, -1001))
sapply(Four_Observation_Groups, function (x) {
  y <- sum(((x - mean(x)) / sd(x)) ^ 5)
})
# [1]  0.000000000  7.500000000  7.312432049 -7.309556249  0.005921752

# Calculating Standardized Sixth Moments for Different Groups of Five
# Observations

Five_Observation_Groups <- list(c(1, 2, 3, 4, 5), c(4, 4, 9, 3, 3), c(10, 100, 1000, 10000, 100000), c(-1, 1, 20, 0, -100), c(-1000, 0, 25, 1000000, 1000001))
sapply(Five_Observation_Groups, function (x) {
  y <- sum(((x - mean(x)) / sd(x)) ^ 6)
})
# [1]  8.320000 29.154365 31.920474 29.807977  3.911114

Answer 1

On top of @whuber's comments, you may also want to consider the minimum number of observations to $\textbf{accurately}$ estimate the moments. As the order $k$ increases, so does the number of observations $n$ you will need to have a reliable estimate. For instance, the variance of the sample skewness will be $\frac{6}{n}$, while the variance of the sample kurtosis will be $\frac{24}{n}$ (under normality assumptions). Although both estimates' variances will decrease at the same rate with $n$, you will need a higher $n$ to reach a given accuracy for a higher moment. See also this related question: Why are larger samples required to estimate higher moments than when estimating the mean?