Central Limit Theorem for difference of two sample means

Question

According to Walpole's Probability and Statistics for Scientists and Engineers, the "central limit theorem can be easily extended to the two-sample, two-population case." That is, if independent samples of size $n_1$ and $n_2$ are drawn at random from two populations with mean $\mu_1$ and $\mu_2$ and variances $\sigma_1^2$ and $\sigma_2^2$, respectively, then the random variable $$Z=\frac{\bar{X}_1-\bar{X}_2-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$$ has approximately the standard normal distribution. My question is, does it require a separate proof, or does it follow from the original version of central limit theorem as a corollary?

Answer 1

The central limit theorem tells you that $\dfrac{\overline X_i - \mu_i}{\sigma_i/\sqrt{n_i}}$ is approximately normally distributed for $i=1,2.$

Without the central limit theorem you know that this random variable has expected value $0$ and variance $1.$

If $\operatorname{var}(\overline X_i-\mu_i) = \sigma_i^2/n_i$ for $i=1,2,$ then $\operatorname{var}\big((\overline X_1 - \overline X_2) - (\mu_1 - \mu_2)\big) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}$ (and you don't need C.L.T. for that).

And the sum of two independent approximately normally distributed random variables is approximately normally distributed.