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Description

Let $N\in\mathbb{N}^{+}$ and $X_{n}\stackrel{IID}{\sim}U(0,1)$ for $n\in\{1,...,N\}$.

Given $X_{1}\leq X_{2}\leq X_{3}\leq...\leq X_{N}$, I would like to understand $f_{X_{n}}$ by writing out $f_{X_{1}}$, $f_{X_{2}}$ and $f_{X_{3}}$ (as part of choosing a proper prior for some Bayesian test).

Attempt

Given $N=2$, by letting $x_{1}=0\implies P(X_{1}\leq X_{2})=1$ such that $f_{X_{2}|X_{1}\leq X_{2}}=f_{X_{2}}=1$ and $x_{1}=1\implies P(X_{1}\leq X_{2})=0$ such that $f_{X_{2}|X_{1}\leq X_{2}}=f_{X_{2}}=0$, my educated guess would be $f_{X_{1}|X_{1}\leq X_{2}}=1-x_{1}$ and $f_{X_{2}|X_{1}\leq X_{2}}=x_{2}$ as these dependances are (probably) linear due to the underlying uniform distributions.

It seems this is confirmed by Bayes theorem as follows $$f_{X_{2}|X_{1}}=\frac{f_{X_{1}|X_{2}}f_{X_{2}}}{f_{X_{1}}}=f_{X_{1}|X_{2}}=\int^{x_{2}}_{0}dX_{1}=x_{2}$$ and $$f_{X_{1}|X_{2}}=\frac{f_{X_{2}|X_{1}}f_{X_{1}}}{f_{X_{2}}}=f_{X_{2}|X_{1}}=\int^{1}_{x_{1}}dX_{2}=1-x_{1}$$ However, for $N=3$ I am stuck as I would not know $f_{X_{1},X_{2}}$ in $$f_{X_{3}|X_{1},X_{2}}=\frac{f_{X_{1},X_{2}|X_{3}}f_{X_{3}}}{f_{X_{1},X_{2}}}$$

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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Commented Feb 22, 2024 at 10:42
  • $\begingroup$ Thank you very much, can I accept or upvote your comment? $\endgroup$ Commented Feb 22, 2024 at 11:04
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    $\begingroup$ You can (maybe after you have achieved sufficient reputation) upvote comments, but in this case, your comment is sufficient already :-). $\endgroup$ Commented Feb 22, 2024 at 11:55
  • $\begingroup$ See stats.stackexchange.com/search?q=%22order+statistic%22+uniform. $\endgroup$ Commented Feb 22, 2024 at 15:31

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