Estimation of the second moment and square root of the second moment (not variance and standard deviation) I want to estimate the second moment of a distribution. I know the breakdown of the second moment into the mean-squared and variance: $\mathbb{E}[X^2] = (\mathbb{E}[X])^2 + var(X)$.
When I want to estimate the second moment, however, it seems dubious to say $\widehat{\mathbb{E}[X^2]} = (\widehat{\mathbb{E}[X]})^2 + \widehat{var(X)} = \bar{x}^2 + s^2$.
In particular, why not $\widehat{\mathbb{E}[X^2]} = \bar{x}^2 + \widehat{\sigma}_{MLE}^2?$
Both seem like they can be defended as estimators of the second moment, but it is not at all clear to me that they have the nice properties like maximum likelihood and/or unbiasedness that we often like in our estimators.
What are the typical ways of estimating the second moment?
 A: Consider $X \sim \mathsf{Norm}(\mu = 5, \sigma=2),$
Then (according to Wikipedia on normal distribution or your text): $$E(X) = 5,\, Var(X) = 2^2 = 4,\, E(X^2) = \mu^2 + \sigma^2 = 25 + 4 =29.$$
The following simulation in R, shows estimates from a million samples of size $n=10$
from this distribution: $$E(\bar X) = E(A) \approx 5,\, E(S^2) \approx 4,\, E(X^2)=E(Q) = E\left(\frac 1n \sum_{i=1}^n X_i^2\right) \approx 29,$$ where $S^2$ is the unbiased estimate of $\sigma^2.$ With a million samples we can expect 2 or 3 significant digits of accuracy.
set.seed(2020)
m = 10^6;  n = 10
x = rnorm(m*n, 5, 2)
DTA = matrix(x, nrow=m)
a = rowMeans(DTA)
q = rowMeans(DTA^2)
mean(a);  mean(s^2);  mean(q)
[1] 4.999994  # aprx E(A) = 5
[1] 3.998543  # aprx Var(X) = 4
[1] 28.99873  # aprx 25 + 4 = 29

If data are from $\mathsf{Exp}(\mu = 5),$
then $$E(X) = \mu = 5,\, SD(X) = \sigma = \mu = 5,\, Var(X) = \mu^2=25,\, E(X^2) = 2\mu^2 = 50.$$
A simulation, similar to the one above for normal data, is as follows:
set.seed(703)
m = 10^6;  n = 10
x = rexp(m*n, .2)
DTA = matrix(x, nrow=m)
a = rowMeans(DTA)
q = rowMeans(DTA^2)
mean(a); mean(q)
[1] 4.998014    # aprx E(X) = 5
[1] 49.96277    # aprx E(X^2) = 50

