# Range of integration for joint and conditional densities

Did I mess up the range of integration in my solution to the following problem ?

Consider an experiment for which, conditioned on $$\theta,$$ the density of $$X$$ is \begin{align*} f_{\theta}(x) = \frac{2x}{\theta^2},\,\,0 < x< \theta. \end{align*} Suppose the prior density for $$\theta$$ is \begin{align*} \pi(\theta) = 1,\,\,\,0 \leq \theta \leq 1 \end{align*} Find the posterior density of $$\theta,$$ then find $$\mathbb{E}[\theta|X]$$. Do the same for $$X = (X_1,\dots, X_n)$$ where $$X_1,\dots, X_n$$ are i.i.d and have the density above.\

The joint density of $$\theta$$ and $$X$$ is given by \begin{align*} f_{\theta}(x)\pi(\theta) = \frac{2x}{\theta^2},\,\,0 < x< \theta \leq 1. \end{align*} and so the marginal density $$g(x)$$ of $$X$$ is given by \begin{align*} g(x)=\int_{x}^1f_{\theta}(x)\pi(\theta)d\theta &= \int_{x}^1\frac{2x}{\theta^2}d\theta\\ &=2x\int_{x}^1\theta^{-2}d\theta\\ &=2x[-\frac{1}{\theta}]_x^1\\ &= -2(x -1),\,\,\,0 So the posterior density of $$\theta$$ is \begin{align*} f_x(\theta) = \frac{f_{\theta}(x)\pi(\theta)}{g(x)} = \frac{-x}{(x-1)\theta^2}, \,\, x < \theta \leq 1 \end{align*} and \begin{align*} \mathbb{E}[\theta|X]&= \int_{x}^1\frac{-x}{x-1}\theta^{-1}d\theta\\ &=\frac{-x}{x-1}\ln\theta|_x^1\\ &= \frac{x}{x-1}\ln x \end{align*} Now let $$X = (X_1,\dots, X_n)$$ where each $$X_i$$ has the density above. Then the joint density is \begin{align*} f_{\theta}(x)\pi(\theta) = \prod_{i = 1}^n\frac{2x_i}{\theta^2},\,\, 0 < x_{[1]} \leq x_{[n]} < \theta \leq 1 \end{align*} and so the marginal density $$g(x)$$ of $$X$$ is given by \begin{align*} g(x)=\int_{x_{[n]}}^1f_{\theta}(x)\pi(\theta)d\theta &= \int_{x_{[n]}}^1\prod_{i = 1}^n\frac{2x_i}{\theta^2}d\theta\\ &=\prod_{i = 1}^n2x_i\int_{x_{[n]}}^1\theta^{-2}d\theta\\ &=\prod_{i = 1}^n2x_i[-\frac{1}{\theta}]_{x_{[n]}}^1\\ &=\Bigg(\frac{1}{x_{[n]}} -1\Bigg) \prod_{i = 1}^n2x_i,\,\,\,0 and so the posterior density is \begin{align*} f_{x}(\theta) = \Bigg(\prod_{i = 1}^n\frac{2x_i}{\theta^2}\Bigg) \cdot \Bigg( \Bigg(\frac{1}{x_{[n]}} -1\Bigg) \prod_{i = 1}^n2x_i \Bigg)^{-1} \end{align*}

The univariate case seems correct to me. The multivariate case should be as follows: \begin{align*} g(x)=\int_{x_{[n]}}^1f_{\theta}(x)\pi(\theta)d\theta &= \int_{x_{[n]}}^1\prod_{i = 1}^n\left(\frac{2x_i}{\theta^2}\right)d\theta\\ &=\left(\prod_{i = 1}^n2x_i\right)\int_{x_{[n]}}^1\theta^{-2n}d\theta\\ &=\left(\prod_{i = 1}^n2x_i\right)\left[-\frac{1}{(2n-1)\theta^{2n-1}}\right]_{x_{[n]}}^1\\ &=\left(\frac{1}{2n-1}\right)\Bigg(\frac{1}{\left(x_{[n]}\right)^{2n-1}} -1\Bigg) \left(\prod_{i = 1}^n2x_i\right),\,\,\,0
Then, the posterior is \begin{align*} f_{x}(\theta) = \frac{2n-1}{\theta^{2n}} \Bigg(\frac{1}{\left(x_{[n]}\right)^{2n-1}} -1\Bigg)^{-1}, x_{[n]}<\theta\leq 1 \end{align*}