1
$\begingroup$

Scenario

From a population, we randomly sample $n$ individuals, measure the normally distributed random variable $X$ and compute the total variance $V_T$ of $X$ in this sample. Then, we randomly split the sample into $m$ groups of size equal size $\frac{n}{m}$ and compute the between-group variance $V_B$ and average within groups variance $V_W$.

Questions

  • What distribution does $V_B$ follows?
  • If the distribution is hard to compute, what is the expected $V_B$?
  • If the distribution is known and easy, could we use this distribution as our null as a surrogate to the standard F-distribution in an ANOVA?

Edit - Interest

My interest in this question comes from a quick thought on selective breeding. A change in the average mean fitness of a population from one generation to the other is equal to the additive genetic variance for fitness. If for reasons of logistic, we must select not on individuals but on groups, then the remaining genetic variance for fitness is the between population variance for fitness $V_B$ and I wanted to understand how much lower is this statistics in comparison to $V_T$. Such calculation would also help me to figure out what group size, number of groups and total population size to choose to allow even weakly selected variants (s≈0.05) to behave significantly different from completely neutral variants.

$\endgroup$
  • $\begingroup$ How would you compute the distribution of any of these variances? e.g. how would you compute the distribution of $V_T$? If you specified a distribution for $X$ such a thing might be possible, but otherwise it's not clear to me how you're going to get anywhere. And while the expectation may be a good deal simpler (under some assumptions), what good is that to you? $\endgroup$ – Glen_b Jun 13 '16 at 5:53
  • $\begingroup$ I forgot to specify that X is normally distributed. I edited the post to specify my interest in this answer. $\endgroup$ – Remi.b Jun 13 '16 at 9:23
  • $\begingroup$ That implies your groups are a sample from a Normal distribution, too, whose parameters are determined by those of $X$ along with $n/m$. With that observation these questions become straightforward to answer. $\endgroup$ – whuber Jun 13 '16 at 15:25
  • $\begingroup$ I don't understand your comment. The observations are sampled from a normal distribution and then split into groups. In other words, each group is also sampled from a normal distribution. Oh this means the between group variance is just the the variance of the sampling distribution. In other words $V_B = \frac{ \sigma^2}{n}$, where $\sigma$ is the variance of X. Is that right? Thanks the comment helped. Should I (1) make an answer (2) let you make an answer or (3) delete my post? Thank you! $\endgroup$ – Remi.b Jun 13 '16 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.