Suppose we have a random sample $X_1,\dots,X_n$ of $X$, where $X$ has the following pdf:
$$f_{\mu,\sigma}(x)=\left(\pi\cdot\sqrt{(x-\mu)(\mu+\sigma-x)}\right)^{-1}$$ where $x\in(\mu,\mu+\sigma),\mu\in\mathbb{R},\sigma\in\mathbb{R}_+$. I have to find a sufficient statistic for this model by Neyman-Fisher factorization theorem.
However I am having difficulties mainly with the math involved to do so.
First, I began by writing the likelihood function, which by i.i.d. hypothesis is just the product of each individual density value for $x_i$, i.e.:
$$\mathcal{L}(\theta\mid x)=\prod_{i=1}^n \left(\pi\cdot\sqrt{(x_i-\mu)(\mu+\sigma-x_i)}\right)^{-1}.$$
I tried to open up the multiplication inside the squared root, which gives me : $$\mathcal{L}(\theta\mid x)=\frac{1}{\pi^n}\prod_{i=1}^n \left(\sqrt{(2\mu+\sigma)x_i-x_i^2-(\mu^2+\sigma\mu)}\right)^{-1}.$$
Then I tried to find some way to write the terms inside the root as a sum of squares but I had no success. I tried to do this in order to be able to simplify the product. If someone would be kind enough to give me some light on how to find the sufficient statistic for this model by factorization theorem, I would be very grateful.
This distribution also intrigues me as I haven't seen it nor found it anywhere.
This is a certain arcsine distribution.
The parent distribution for $\mu\in\mathbb R$ and $\sigma>0$ is $$f_{\mu,\sigma}(x)=\frac{\mathbf1_{\mu<x<\mu+\sigma}}{\pi\sqrt{(x-\mu)(\mu+\sigma-x)}}$$
So the likelihood given the sample $x_1,\ldots,x_n$ is
$$L(\mu,\sigma\mid x_1,\ldots,x_n)=\frac{\mathbf1_{\mu<x_{(1)},x_{(n)}<\mu+\sigma}}{\pi^n\prod_{i=1}^n\sqrt{(x_i-\mu)(\mu+\sigma-x_i)}}\,,$$
where $x_{(1)}=\min\limits_{1\le i\le n}x_i$ and $x_{(n)}=\max\limits_{1\le i\le n}x_i$ as usual.
There is no non-trivial sufficient statistic using Factorization theorem as we cannot factor out the likelihood in the form $g(T(x_1,\ldots,x_n),\mu,\sigma)h(x_1,\ldots,x_n)$ where the statistic $T$ is free of $(\mu,\sigma)$.
Hence you only have the sample $(X_1,\ldots,X_n)$ and the order statistics $(X_{(1)},\ldots,X_{(n)})$ as your trivial sufficient statistics.