Assume $X_i$ are generated with the following distribution:
$$ f(x; \theta, c) = \theta^{-c}cx^{c-1}e^{-(x/\theta)^c}$$ $\theta>0$ and $c>0$ is known.
Further, assume $T(X)=\sum^{n}_{i=1} X_i^c$. How can we derive the distribution of $T(X)$ such, that it would get the form of something along the lines of $\theta^c\cdot \chi^2(2n)$ distribution?
Also, I have a guess that any distribution, that can be expressed as an exponential family, their statistic $T(x)$ can be also expressed as a $\chi^2$ with some scale/mean transformation. Is this true, and maybe there is some literature describing exactly this?
Assuming $X_1,X_2,\ldots,X_n$ are i.i.d with pdf
$$f(x;\theta,c)=\frac{cx^{c-1}e^{-(x/\theta)^c}}{\theta^c}\mathbf1_{x>0}\quad,\,\theta,c>0$$
It is straightforward to show that $$\left(\frac{X_i}{\theta}\right)^c\stackrel{\text{ i.i.d }}\sim \mathsf{Exp}(1)$$
Equivalently, $$2\left(\frac{X_i}{\theta}\right)^c\stackrel{\text{ i.i.d }}\sim \chi^2_2$$
Hence, $$2\sum_{i=1}^n\left(\frac{X_i}{\theta}\right)^c=\frac{2}{\theta^c}T\sim \chi^2_{2n}$$