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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.

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    $\begingroup$ Suggestion: use the characteristic functions. $\endgroup$
    – whuber
    Sep 4, 2021 at 13:04
  • $\begingroup$ @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 ? $\endgroup$
    – Fiodor1234
    Sep 4, 2021 at 13:24
  • $\begingroup$ Oh nevermind I think I got how to solve it $\endgroup$
    – Fiodor1234
    Sep 4, 2021 at 13:29

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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.

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

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    $\begingroup$ 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. $\endgroup$
    – whuber
    Sep 5, 2021 at 15:08
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    $\begingroup$ @whuber thank you very much for your comment, I was really looking for a clarification there $\endgroup$
    – Fiodor1234
    Sep 5, 2021 at 15:17
  • $\begingroup$ @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 $\endgroup$
    – Fiodor1234
    Sep 5, 2021 at 15:25
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    $\begingroup$ Thank you for you guys! I got to know that the Inversion theorem on this occasion! And that this theorem is really helpful! $\endgroup$
    – Minho Kang
    Sep 6, 2021 at 7:43
  • $\begingroup$ @MinhoKang The Inversion Theorem, gives you the reasoning that you can use characteristic functions to derive your result $\endgroup$
    – Fiodor1234
    Sep 6, 2021 at 10:11

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