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Let $X_1, \ldots, X_n$ iid from a $Unif(0,\theta)$ distribution, and let $X_{(n)}$ be the $n$th order statistic, that is, the maximum of $X_1, \ldots, X_n$. I would like to find $E(X_1|X_{(n)})$. I have that when $X_{(n)}=t$, $X_1=t$ with probability $\frac{1}{n}$, and that with probability $\frac{n-1}{n}$, $X_1$ appears to be uniform on $(0,t)$ (not sure why). I am unable to translate these into a functional form. It seems like:

\begin{align} E(X_1|X_{(n)}=t) &= E(X_1=t|X_{(n)}=t)P(X_1=t)+E(X_1\in (0,t)|X_{(n)}=t)P(X_1\in (0,t))\\ &= t\cdot\frac{1}{n}+ \int_0^{t}x\ f(x|X_{(n)}=t)dx \end{align}

But I am unsure why it is that with probability $\frac{n-1}{n}$, $X_1$ appears to be uniform on $(0,t)$ and if the integral approach even makes sense?

user321627
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    Intuitively, $X_{(n)}=X_1$ with probability $\frac1n$ as each observation is equally likely to be the largest; the probability of two equally largest values is zero for a continuous distribution. Otherwise, so with probability $1-\frac1n$, you have $X_{(n)} \gt X_1$ and given the value of $X_{(n)}$ you have $X_1$ conditionally uniformly distributed between $0$ and the value of $X_{(n)}$. – Henry Dec 07 '16 at 12:05
  • Also see https://math.stackexchange.com/questions/1007744/conditional-expectation-to-de-maximum-ex-1-mid-x-n, https://math.stackexchange.com/questions/617415/compute-ex-1y?noredirect=1&lq=1. – StubbornAtom Sep 18 '19 at 13:59

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