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Asking for help in approaching a question from a Statistics textbook:

Let $X_1, X_2, ..., X_n$ independent and identically distributed with density function $f_ {\theta}(x)$ and $T_n(X_1,X_2,...X_n)$ a sufficient statistic for $\theta$ that is a continuous random variable. Let $g_{\theta}(t)=F_{T_n,\theta}(t)=\mathbb{P}_{\theta}(T_n\leq t)$ be the cumulative distribution function of $T_n$, and $Y_n=g(T_n)$. Show that $Y_n\sim U(0,1)$ and thus $Y_n$ is a pivot.

My initial thoughts were to somehow find the distribution of $T_n$ using a change of variables, but $T_n$ is not the same dimension as the probability space of the sample $X_1,..X_n$. Or to somehow use the sufficiency to write $f_{\theta}(x)=h(x)g_{\theta}(T_n(x))$, not sure though how to proceed from here.

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    $\begingroup$ The beginning is useless. As soon as $T_n$ is a continuous random variable, and $g$ is its cumulative distribution function, the composition $g(T_n)$ is uniform over $(0,1)$. $\endgroup$ Commented Oct 24 at 7:05
  • $\begingroup$ see e,g, chatgpt.com/s/t_68fb26d41430819191b10be64141e56c for the proof.... $\endgroup$ Commented Oct 24 at 7:12
  • $\begingroup$ You seem to need $T_n$ to be one dimensional for $\mathbb{P}_{\theta}(T_n\leq t)$ to make sense. $\endgroup$ Commented Oct 24 at 9:01

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Thanks for the comments. $Y_n\sim U(0,1)$ is a consequence of the probability integral transform, and $T$ being a sufficienct statistic is irrelevant - only that it's a continuous variable.

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