Distribution of the mean of samples taken from two different distributions

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 $$\{4,5,6\}$$

• 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 "$n$"?
– whuber
Commented Jul 24, 2023 at 13:11
• $n_1, n_2$ 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 Commented Jul 24, 2023 at 14:23
• 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.
– whuber
Commented Jul 24, 2023 at 14:34

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 $$n_2/n\to 1$$