I am currently reading about the standard error. I know about central limit theorem, but I don't understand why, if my variable is normally distributed in the population, the sampling distribution (the distribution of the sample means) is also a normal distribution, no matter of sample size etc.
Why is the sampling distribution of normal distributed variable automatically also normal distributed
This is not an asymptotic result, it is a finite-sample result and has nothing to do with the Central Limit Theorem.
You start by assuming that the population is normal. The normal distribution is one of the few distributions that are closed under addition, meaning that the sum of normal random variables follows a normal distribution (although with different parameters than the components of the sum). More over, scaling a normal random variable, leaves it a normal random variable (again, affecting the parametrization of it).
The sample mean of a sample from a normal population, is a sum of normal random variables, scaled by $1/n$. Hence it too follows a normal distribution.
The Central Limit Theorem comes into effect to lead to the same result asymptotically, when the population is not normally distributed.
3$\begingroup$ +1 Because statistical language can be confusing, especially to beginners, I would like to point out that scaling does change the distribution: but it does not change its shape (which remains Normal) and that of course is what you are contending. $\endgroup$– whuber ♦Apr 16, 2014 at 19:26
$\begingroup$ So very true. Corrected. $\endgroup$ Apr 16, 2014 at 19:54
$\begingroup$ +1 Some people phrase it as 'scaling doesn't change the family of distributions', because the normal isn't so much a distribution as a location-scale family of distributions. (Don't take this as a suggested change to the answer, though, which seems fine as it is.) $\endgroup$– Glen_bApr 16, 2014 at 22:18
$\begingroup$ @Glen_b Yes I have noticed that some times what is called "distribution" in one corner, is called "family of distributions" in another. I have the impression that the latter is used more by those that are more deeply into statistics. $\endgroup$ Apr 16, 2014 at 22:35
1$\begingroup$ @MSIS Either that they are independent, or that they are dependent but follow jointly a multivariate Normal. The standard example is $X$ and $Y = -X$. If $X$ is Normal, so is $Y$, but $X +Y = 0$, a degenerate r.v. Moreover $(X,Y)$ do not follow jointly a multivariate Normal. $\endgroup$ Dec 19, 2022 at 2:05