@whuber Thank you! I tried to simulate data by code: x = list() for mu in np.random.normal(0, 1/sqrt(2), 10000): sigma = 1/sqrt(2) x.extend(list(np.random.normal(mu, sigma, 10))) And I got a perfect matching with np.random.normal(0, 1, 100000).

@jginestet I mean, if I select a set of distributions $N(\mu_1, \sigma^2/2)$, $N(\mu_2, \sigma^2/2)$, ..., $N(\mu_n, \sigma^2/2)$ from $N(0, \sigma^2)$ and combine them into one set (all samples from selected distributions), then I will get the original distribution $N(0, \sigma^2)$. The question is: what is the distribution for $\mu$ should be?

@Dave Thank you! I have $N(0, \sigma^2)$, from which I can select $N(\mu, \sigma^2/2)$. In other words, if there are a set of $N(\mu, \sigma^2/2)$ distributions that together form $N(0, \sigma^2)$ distribution, what the distribution for $\mu$ should be? There are two extreme points. This is if I chose distributions with the same standard deviation $\sigma^2$. In this case, the standard deviation for $\mu$ would tend to zero with n->inf. And if I were to select one point from $N(0, \sigma^2)$, then the standard deviation for $\mu$ would be equal to $\sigma^2$.