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When testing for a difference in the means of two independent normal samples with assumed equal variance $\sigma^2$, we use the statistic $T=\frac{\overline{X_1}-\overline{X_2}}{S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, which can equivalently be expressed as $\frac{\left(\frac{\overline{X_1}-\overline{X_2}}{\sigma\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \right)}{\sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{\sigma^2(n_1+n_2-2)}}}$, where $S_1,S_2$ represent the sample standard deviations of the two samples, and $n_1, n_2$ represent the sample sizes.

I've been reading proofs as to why T follows a t-distribution, and they all use the fact that $\frac{Z}{\sqrt \frac{\chi^2_n}{n}} \sim t_n$, given that $Z$ and $\chi^2_n$ are independent, where Z is a standard normal distribution. I can see why Z=$\left(\frac{\overline{X_1}-\overline{X_2}}{\sigma\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \right)$ follows a standard normal distribution, and why $Y^2 = \frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{\sigma^2}$ follows a chi square distribution with $n_1+n_2-2$ degrees of freedom. But, I've never seen an argument that shows why $Z$ and $Y^2$ are independent.

If someone could either post the proof here, or redirect me to some source with the proof, it would be greatly appreciated.

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$\bar{X_1}$ and $S^2_1$ are independent. $\bar{X_2}$ and $S^2_2$ are independent. This independence of $\bar{X}$ and $S^2$ is true because of normality condition. proof-of-the-independence-of-the-sample-mean-and-sample-variance .

Two samples are independent.

If $X$ and $Y$ are independent so any (measurable) function of them ( $g(X)$ and $g(Y)$)are independent. if-x-and-y-are-independent-then-fx-and-gy-are-also-independent. So

$$P(\bar{X_1}\in A , \bar{X_2}\in B ,S^2_1 \in C,S^2_2 \in D) =P(\bar{X_1}\in A,S^2_1 \in C)*P(\bar{X_2}\in B,S^2_2\in D)$$ since Two samples are independent , so any function of them are independent. $$=P(\bar{X_1}\in A) P(S^2_1 \in C)P(\bar{X_2}\in B)P(S^2_2\in D)$$ Since $\bar{X}$ and $S^2$ are independent.

You can see it clearly by $$P(\bar{X_1}\in A , \bar{X_2}\in B ,S^2_1 \in C,S^2_2 \in D) =P(\bar{X_1}\in A) P(S^2_1 \in C)P(\bar{X_2}\in B)P(S^2_2\in D) =P(\bar{X_1}\in A,\bar{X_2}\in B)*P(S^2_1 \in C, S^2_2\in D)$$

so $(\bar{X_1},\bar{X_2})$ and $(S^2_1 , S^2_2)$ are independent so any function of them($\bar{X_1}-\bar{X_2}$ and $g(S^2_1 , S^2_2)$) are independent.

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  • $\begingroup$ Yep I was already aware of this fact - however as far as I know the following statement isn't necessarily true: "if X, Y, Z, W are independent, then X-Y is independent of any weighted sum of Z and W". Unless you are saying that it does hold true specifically in the case of normal random variables? $\endgroup$ – user767761 Apr 7 '20 at 10:48
  • $\begingroup$ $(\bar{X_1},\bar{X_2})$ and $(S^2_1 , S^2_2)$ are independent so any function of them($\bar{X_1}-\bar{X_2}$ and $g(S^2_1 , S^2_2)$) are independent. $\endgroup$ – Masoud Apr 7 '20 at 11:14

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