Show that noncentral $X_2$ is $\chi^2(r-r_1,\theta -\theta_1)$ As I read a book named 'Introduction to mathematical statistics' written by Hogg et al, I stuck with the below question.
$X_1$ and $X_2$ be two independent random variables. And Let $X_1$and $Y=X_1+X_2$ be $\chi^2(r_1,\theta_1) $ and $\chi^2(r,\theta)$, respectively. Here $r_1\le r$ and $\theta_1\le\theta$.
Could you explain that $X_2$ is $\chi^2(r-r_1,\theta -\theta_1)$?
This question looks so simple, but I'm not good at statistics so I want to know that. Really Thank you for hands.

There is the Inversion Theorem, which states that a probability density function can be derived from the Characteristic function, here are two pdf that discusses this matter with proofs
https://sas.uwaterloo.ca/~dlmcleis/s901/chapt6.pdf, https://nptel.ac.in/content/storage2/courses/108106083/lecture26_CF.pdf
The important is that if $C_{X}(t)$ corresponds to the characteristic function of a random variable $X$, then you can express the density function in terms of $C_{X}(t)$ as:
$$f_{X}(x)=\frac{1}{2\pi}lim_{T\rightarrow \infty} \int_{-T}^{T}e^{-itx}C_{X}(t)dt$$
So, if you prove that $\mathbb{E}[e^{itX_{2}}]= \frac{e^{\frac{i(r-r_{1})t}{(1-2it)}}}{(1-2it)^{(\theta-\theta_{1})/2}}$
Then you will know that $X_{2}$ follows your desired distribution
Also, note that if $A$ and $B$ are two independent random variables then the characteristic function of their sum can be decomposed, i.e.
$$\mathbb{E}[e^{it(a+b)}]=\mathbb{E}[e^{ita}]\mathbb{E}[e^{itb}]$$
