Monte Carlo simulation or big samples

Question

Suppose that we have a population that behaves as $N(1,\sigma^2)$, and we want to estimate $\sigma^2$. I would like to understand the difference between these two approaches:

1. We take a sample $x_1,\ldots,x_m$ for large $m$ and estimate $$\hat{\sigma}^2=\frac{\sum_{i=1}^m (x_i-1)^2}{m-1}=:T(x_1,\ldots,x_m).$$

2. We perform Monte-Carlo simulation: for large $n$ and $N$, take vectors of samples, $$\vec{x}^1=(x_1^1,\ldots,x_n^1),\ldots,\vec{x}^N=(x_1^N,\ldots,x_n^N),$$ and estimate $$\hat{\sigma}^2=\frac{1}{N}\sum_{j=1}^N T(\vec{x}^j).$$

If $nN=m$, we have the same number of observations, so what are the differences?

EDIT: What I want to know is the difference between taking big samples (as in 1) or Monte Carlo simulation (as in 2). Do not focus on the particular case of the normal, since I just wrote an (unreal) example. I would like an answer applicable to any distribution.

Answer 1

Imagine that you want to estimate mean age of some population, so you draw some large sample from the population, collect the data about age of the study participants and calculate mean from it.

If you conducted simulation you would assume some distribution for age and assume parameters (or hyperparameters), then you would conduct the simulation and calculate the mean of the values created using the simulation. Mean of the simulation is mean of the simulation, it doesn't have to have anything in common with anything appearing in the real world unless you conduct your simulation in such manner so that the simulation resembles the real world phenomenon.

For example, you could conduct the simulation of human age using a uniform distribution bounded in $[0,1]$ and then conclude that the average age is $0.5$, what obviously would have nothing to do with age of humans in the real world.

Answer 2

Answer for normal distribution using small number correction which is needed for unbiased standard deviation:

$\text{E}(s)\,=\,\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{2}}>\text{SD}=\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}\; .$

Thus, $\frac{E(s)}{SD}=\frac{\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{2}}}{\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}}=\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}\sqrt{\frac{n-1}{2}}$.

Thus, for a sample size of $m>N$, the correction is lesser (substitute $m$ for $n$ in the above correction) than for $N$ (substitute $N$ for $n$ in the above correction), unless $N>>100$. This comes from Why is sample standard deviation a biased estimator of $\sigma$? For non-normal distributions a similar derivation for each one would be in order.