1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ If the samples are independent then the central limit theorem is not necesary. Also if you accept that the sum/difference of two normal distributed variables are normal distributed as well then the mean and the variance follows straightforward. $\endgroup$ – callculus Oct 20 '20 at 3:21
  • 1
    $\begingroup$ @callculus : Why would you say that independence means the C.L.T. is not needed? The populations were not assumed to be normally distributed, so the C.L.T. is the only reason to think the sample means are approximately normally distributed. $\endgroup$ – Michael Hardy Oct 20 '20 at 3:46
  • 1
    $\begingroup$ @MichaelHardy I missed that. Then the Central Limit Theorem is necessary. Thanks that you point that out. $\endgroup$ – callculus Oct 20 '20 at 3:54
2
$\begingroup$

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.

$\endgroup$

Your Answer

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

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