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I have a question regarding the use of Basu's Lemma. Assume that $$X \sim N(\theta, \sigma^2).$$ In that case, we can use Basu's Lemma to show that $$\bar{X}$$ and $$S^2= \frac{1}{n-1}\sum(X_i-\bar{X})^2$$ are independent, if we show that $T:=\bar{X}$ is complete and sufficient (e.g. by the definition of exponential families) and that the distribution of $S^2$ does not depend on $\theta$ which is the case here. However, when $\sigma^2$ would be unknown as well, the distribution of $S^2$ would depend on $\theta$. Can we still use Basu's Lemma somehow to show independence in that case ? I was confused by the following statement in the book Statistical Inference by Casella and Berger on p.289:

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StubbornAtom
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Vala
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    What makes you think "when $σ^2$ would be unknown as well, the distribution of $S^2$ would depend on $θ$"? – Henry Jan 10 '20 at 10:07
  • @Henry I understand the question as follows: what if $\sigma^2$ depends on $\theta$? Say, $\sigma^2=\theta^2$, $\theta\neq 0$. Is Basu's Lemma applicable here? – NCh Jan 10 '20 at 11:32
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    @NCh I would read the question as having $\theta$ and $\sigma^2$ unknown but fixed – Henry Jan 10 '20 at 11:38
  • @Henry For $N(\theta,\theta^2)$ parameter $\theta$ is also unknown but fixed. I never supposed bayesian settings. – NCh Jan 10 '20 at 11:40
  • @Henry I cannot answer the OP question, because, following local traditions in theoretical statistics, I am used to proving the independence of these quantities in lectures without using Basu's lemma. And no collisions with sufficiency and completeness (as in this example) arise. – NCh Jan 10 '20 at 11:50
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    When both $\theta$ and $\sigma^2$ are unknown, distribution of $S^2$ depends on $\sigma^2$ only; distribution $\overline X$ depends on $\theta$ in any case. So neither the distribution of $\overline X$ nor that of $S^2$ is independent of the parameter $(\theta,\sigma^2)$ as mentioned in highlighted sentence. (BTW I don't understand why this is flagged as 'belongs on stats.SE'.) – StubbornAtom Jan 10 '20 at 15:27
  • And indeed we can still use Basu's theorem. See example 6.6.15 here or https://math.stackexchange.com/a/50598/321264. – StubbornAtom Jan 10 '20 at 15:30
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