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Suppose we are given that $X_1$ and $X_2$ are two non-negative random variable such that $$X_1+X_2\sim\chi^2_{(2)}$$. Also we are given that $X_1\sim\chi^2_{(1)}$. Can we say the following

  1. $X_2$ is independent of $X_1$?
  2. $X_2\sim\chi^2_{(1)}$?

If not, then please provide a counterexample.

This is just a fun question question. I'm asking it just for curiosity. Any kind of help appreciated.

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    $\begingroup$ No - it is not difficult to have $X_2$ a deterministic function of $X_1$. What have you tried? $\endgroup$
    – Henry
    Oct 16, 2022 at 17:11
  • $\begingroup$ Here is a Python gist that (unsuccessfully) attempts to synthesize an approximate counterexample. More tweaking is needed to favor getting distribution shapes approximately correct. $\endgroup$
    – Galen
    Oct 16, 2022 at 17:54
  • $\begingroup$ @Henry I tried using $(X_1+X_2, X_1)=A\cdot(X_1,X_2)$ where $A$ is a matrix but that doesn't help. $\endgroup$
    – annie_lee
    Oct 17, 2022 at 3:56

2 Answers 2

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Suppose $F_{(1)}(x)$ is the cumulative distribution function of $\chi^2_{(1)}$, and $F^{-1}_{(1)}(x)$ its inverse i.e. quantile function, and $F_{(2)}(x)$ is the cumulative distribution function of $\chi^2_{(2)}$.

Then

  • if $X_1 \sim \chi^2_{(1)}$ so $F_{(1)}(X_1)\sim Unif(0,1)$
  • and $X_2 =F^{-1}_{(2)}\left(F_{(1)}(X_1)\right)-X_1$ so $X_2$ is determined by $X_1$ and thus not independent
  • then $X_3 =X_1+X_2 \sim \chi^2_{(2)}$

$X_2 \not \sim \chi^2_{(1)}$: for example it has a variance of about $0.409$ while $X_1$ has a variance of $2$ and $X_3$ has a variance of $4$

requested in comments:

This $X_2$ is non-negative because you have $F_{(1)}(x) \ge F_{(2)}(x)$ and then apply a non-decreasing function gives $F^{-1}_{(2)}\left(F_{(1)}(x)\right) \ge x$ and thus $X_2 =F^{-1}_{(2)}\left(F_{(1)}(X_1)\right)-X_1\ge 0$

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  • $\begingroup$ Isn't it $X_2= F_2^{-1}(F_1(X_1)) - X_1$?and $F_1(X_1)\sim Unif[0,1]$. $\endgroup$
    – annie_lee
    Oct 17, 2022 at 12:28
  • $\begingroup$ @annie_lee Good point - edited - thank you $\endgroup$
    – Henry
    Oct 17, 2022 at 12:55
  • $\begingroup$ Could you explain why $X_2$ is also a nonnegative random variable as required by the hypotheses? $\endgroup$
    – Dilip Sarwate
    Oct 17, 2022 at 16:33
  • $\begingroup$ @DilipSarwate You have $F_{(1)}(x) \ge F_{(2)}(x)$ and so apply a non-decreasing function $F^{-1}_{(2)}\left(F_{(1)}(x)\right)\ge x$ and thus $X_2 = F^{-1}_{(2)}\left(F_{(1)}(X_1)\right)-X_1 \ge 0$ $\endgroup$
    – Henry
    Oct 17, 2022 at 16:45
  • $\begingroup$ Please do take the time to edit your answer to incorporate your response into it. Thanks. $\endgroup$
    – Dilip Sarwate
    Oct 17, 2022 at 16:51
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Another way to construct a counterexample: let $Z\sim\chi^2_2$ be independent of $X_1$, and define $X_2=Z-X_1$. Since $X_2$ is not even strictly non-negative, it's not $\chi^2_1$.

However, if $X_2$ is independent of $X_1$, then $X_2\sim \chi^2_1$: the characteristic function of $X_1+X_2$ is the product of the two individual characteristic functions, which is enough to determine the distribution of $X_2$.

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  • $\begingroup$ Your $X_2$ would have variance $6$ here, more evidence that it is not $\chi^2_1$ $\endgroup$
    – Henry
    Oct 17, 2022 at 12:59
  • $\begingroup$ The hypotheses are that both $X_1$ and $X_2$ are nonnegative random variables. It is true that $X_2$ is not $\chi_1^2$ but so what? The OP wants $X_2$ to be nonnegative, just like $X_1$ is, $\endgroup$
    – Dilip Sarwate
    Oct 17, 2022 at 16:42

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