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I'm having trouble with this question.

Let $X_1, \ldots, X_n$ be a random sample from some distribution with mean $\mu_1$ and variance $\sigma_1^2>0$. Let $Y_1, \ldots, Y_n$ be a random sample from some other completely unrelated distribution with mean $\mu_2$ and variance $\sigma_2^2>0$. Prove that as $n \to \infty$, $$ Z_n = \frac{(\bar{X}_n - \bar{Y}_n) - (\mu_1-\mu_2)} {\sqrt{\left(\sigma_1^2 + \sigma_2^2 \right)}/n} \to \mathcal{N}(0,1). $$

The fact that I don't know the distribution of $X_i$ and $Y_i$ throws me off. Does the CLT always go to $\mathcal{N}(0,1)$?

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If you define $A_n = X_n - Y_n$ then you don't know the distribution of $A_n$ but its mean is $\mu_A = \mu_1 - \mu_2$ and its variance is $\sigma_A^2 = \sigma_1^2 + \sigma_2^2$ since $X$ and $Y$ are unrelated (i.e. independent) processes.

Hence, $$ Z_n = \frac{\frac{1}{n}\sum_{k=1}^n A_n - \mu_A}{\sigma_A/n} $$

and you can apply the CLT.

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  • $\begingroup$ Does the CLT always have a distribution of $\mathcal{N}(0,1)$?, so since we can apply the CLT, we can conclude that this distribution is also $\mathcal{N}(0,1)$? $\endgroup$ – user270494 Feb 18 '16 at 19:45
  • $\begingroup$ @user270494 depends on the form. In the most common form for a stochastic process $Z_n$ with mean $m$ and stdev $s$, we have $$\frac{\bar{Z}_n - m}{s/n} \to \mathcal{N}(0,1)$$ $\endgroup$ – gt6989b Feb 18 '16 at 19:48
  • $\begingroup$ So since we can write it in that form, then $Z_n$ -> $\mathcal{N}(0,1)$ $\endgroup$ – user270494 Feb 18 '16 at 20:06
  • $\begingroup$ @user270494 yes $\endgroup$ – gt6989b Feb 19 '16 at 1:52

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