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 $\{4,5,6\}$

  • $\begingroup$ 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


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 $n_2/n\to 1$


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