Expectation of a statistic conditional on another statistic

Question

$X_1,\ldots,X_n$ are i.i.d with density $f(x;\theta) = \theta x^{\theta - 1}, 0<x<1$

I'm trying to work out:

$$\mathbb{E}(-\log X_1 \mid \sum_{i=1}^n \log X_i = k)$$

How do I go about this? I tried rearranging the stuff inside the expectation to give

$$\mathbb{E}(-\log X_1 \mid \sum_{i=1}^n \log X_i = k) = \mathbb{E}(-\log X_1 \mid -\log X_1 = \sum_{i=2}^n \log X_i - k)$$

$$= \mathbb{E}(\sum_{i=2}^n \log X_i - k) = \sum_{i=2}^n \mathbb{E}(\log X_i) - k$$

but I don't think this is correct. Thanks

Answer 1

$\newcommand{\E}{\operatorname{E}}$ \begin{align} & \E (-\log X_1 \mid \sum_{i=1}^n \log X_i = k) + \cdots + \E(-\log X_n \mid \sum_{i=1}^n \log X_i = k) \\[10pt] = {} & \E\left( -\log X_1-\cdots-\log X_n \mid \sum_{i=1}^n \log X_i = k \right) \\[10pt] = {} & \E\left( -\sum_{i=1}^n \log X_i \mid \sum_{i=1}^n \log X_i = k \right) = -k. \end{align}

If you can prove that all $n$ of the terms in the first line above are equal to each other, then each must be $-k/n$.