0
$\begingroup$

Consider you have some distributions $Z_1$ and $Z_2$ of which you select $n_1$ samples from $Z_1$ and $n_2$ samples from $Z_2$.

We now add up these samples and take the mean. The question becomes, what is the distribution of $\mu$? Can this distribution be approximated via the CLT for sufficiently large values if $n$?

Normally this would be a standard CLT question if it involved just one distribution, but I am not sure how two different distributions with different sample sizes can effect this.


In case you want more specific examples of the problem, $Z_1$ and $Z_2$ are both uniform distributions on integers, just different ranges of integers. For example, $Z_1$ can be uniform selection of $\{1,2,3\}$ and $Z_2$ uniform selection of $\{4,5,6\}$

$\endgroup$
3
  • $\begingroup$ When $n_1$ has a Binomial$(n,p)$ distribution for some $p,$ this is equivalent to sampling from a mixture of two distributions and so the CLT applies directly. Otherwise, what can you tell us about how the $n_i$ are related to your "$n$"? $\endgroup$
    – whuber
    Commented Jul 24, 2023 at 13:11
  • $\begingroup$ $n_1, n_2$ are pre-choosen per some proportion. For example $.9$ samples are chosen from $Z_1$ and $.1$ samples are chosen from $Z_2$. My goal is to see how varying this proportion impacts the distribution $\endgroup$
    – wjmccann
    Commented Jul 24, 2023 at 14:23
  • $\begingroup$ Why invoke the CLT, then? Why not just examine the distributions? This question could have an exact and even an analytical answer, if the distributions are given in that form; and otherwise could easily be determined through numerical means. $\endgroup$
    – whuber
    Commented Jul 24, 2023 at 14:34

1 Answer 1

1
$\begingroup$

This can be done using a more general CLT such as the Lindeberg-Feller or Lyapunov CLT, which don't assume the same distribution (just independence and a tail condition). In particular, the Lindeberg-Feller CLT gives necessary conditions for an average of independent observations to be asymptotically Normal.

It can also be done directly using the classical CLT: we know that $$\sqrt{n_1}(\bar Z_{1n_1}-\mu_1)\stackrel{d}{\to} N(0,\sigma^2_1)$$ where $\mu_1$ and $\sigma^2_1$ are the mean and variance of $Z_1$ and $\bar Z_{1n_1}$ is the average. In the same way $$\sqrt{n_2}(\bar Z_{2n_2}-\mu_2)\stackrel{d}{\to} N(0,\sigma^2_2).$$

We can write the overall average $\bar Z_n$ as $$\bar Z_n = \frac{n_1}{n}\bar Z_{1n_1}+\frac{n_2}{n}\bar Z_{2n_2}$$

Now we can either make assumptions about $n_1/n$ having a limit or, more generally, work along subsequences where it has a limit (any subsequence has a subsubsequence where there's a limit)

Along subsequences where $n_1/n\to p_1\in(0,1)$, we can define $\mu_n=p\mu_1+(1-p)\mu_2$ and $\sigma^2=p^2\sigma^2_1+(1-p)^2\sigma^2_2$ $$\sqrt{n}(\bar Z_n-\mu_n)\stackrel{d}{\to} N(0,\sigma^2).$$

Along subsequences where $n_1/n\to 1$, there asymptotically isn't any $Z_2$ contribution, so $$\sqrt{n}(\bar Z-\mu_1)\stackrel{d}{\to} N(0,\sigma^2_1)$$ and the other way around when $n_2/n\to 1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.