# Transforming sum of exponential variables to chi-squared distribution

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, $$2\sum_{i=1}^n\left(\frac{X_i}{\theta}\right)^c=\frac{2}{\theta^c}T\sim \chi^2_{2n}$$