# 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.

• Suggestion: use the characteristic functions.
– whuber
Sep 4, 2021 at 13:04
• @whuber when you use characteristic function you will have to calculate $\mathbb{E}[e^{itX_{2}}] = \mathbb{E}[e^{itY}e^{i(-t)X_{1}}]$, however $X_{1}$ and $Y$ are not independent how do we proceed from there ? Sep 4, 2021 at 13:24
• Oh nevermind I think I got how to solve it Sep 4, 2021 at 13:29

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.

$$\mathbb{E}[e^{it(a+b)}]=\mathbb{E}[e^{ita}]\mathbb{E}[e^{itb}]$$

• You misquote the Portmanteau Lemma: the convergence has to hold for every bounded continuous function, not just one of them! A different theorem is needed here: namely, that the characteristic function determines the distribution.
– whuber
Sep 5, 2021 at 15:08
• @whuber thank you very much for your comment, I was really looking for a clarification there Sep 5, 2021 at 15:17
• @whuber Eventually the result that if two characteristic functions agree then their random variables have the same distribution is a consequence of a Theorem called Inversion Formula Sep 5, 2021 at 15:25
• Thank you for you guys! I got to know that the Inversion theorem on this occasion! And that this theorem is really helpful! Sep 6, 2021 at 7:43
• @MinhoKang The Inversion Theorem, gives you the reasoning that you can use characteristic functions to derive your result Sep 6, 2021 at 10:11