# What is the pdf of sample mean and variance, given the population mean and variance?

Suppose I have drawn n samples from a population of known mean and variance ( for example, a normal distribution with mean zero and variance 1.0 ).

I then calculate the mean and standard deviation of the sample.

How do I calculate the pdf of these sample values, given that I know the population values?

• The procedure is illustrated at stats.stackexchange.com/questions/68984, as well as in many more of the thousands of posts found by searching sampling distribution. Most of them focus on the sample mean; a few on the sample correlation; and probably none of them describe how to compute the sampling distribution of the SD because (with few exceptions) it is an intractable calculation.
– whuber
Feb 23, 2014 at 16:25
• The variance is generally substantially more straightforward than the standard deviation. Feb 23, 2014 at 16:38

The distributions of the sample mean and variance of a normal distribution are well-known (normal for the mean, Chi square for the variance). As whuber says, you can't find the pdfs of the sample mean $\overline{x}$ and, especially, the variance $s^2$ except in special situations. Given only the population mean $\mu$ and variance $\sigma^2$ and nothing else, all you can find exactly are sample mean and variance of $\overline{x}$ and the mean of $s^2$ (but not $s$):

Let the sample size be $n$. Then every introductory text on statistical theory demonstrates that:

$$E(\overline{x})= \mu$$

$$Var(\overline{x}) = \frac{\sigma^2}{n}$$ and $$E(s^2) = \sigma^2$$

If you know, in addition to $\mu$ and $\sigma^2$, the population fourth central moment $\mu_4 = E[(X =\mu)^4]$, you can also compute the exact variance of $s^2$

$$Var(s^2) = \frac{(n-1)^2}{n^3}\left(\mu_4 - \frac{n-3}{n-1}\sigma^2\right)$$

Reference:

CR Rao (1965) Linear Statistical inference and its applications, Wiley, New York, p.368.