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I have gotten to the point where I found the joint pdf of $Z_1$ and $Z_2$ is

$h_{1,2}(z_1,z_2) = 2f(z_1)f(z_1+z_2)$,

but I do not know where to go from here because when I assume the independence of $Z_1$ and $Z_2$, I cannot get anywhere:

$h_{1,2}(z_1,z_2) = 2f(z_1)f(z_1+z_2) = h_1(z_1)h_2(z_2)$,

where $h_1(z_1)$ and $h_2(z_2)$ represent the marginal pdf of $Z_1$ and $Z_2$, respectively.

Some_Math_Nerd
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    This characterization of the exponential distribution is discussed in https://www.jstor.org/stable/2238249?seq=1. – StubbornAtom Oct 01 '20 at 07:42
  • I have looked at the research article and they show that if $Y_1$ and $Y_2$ are independent then in order for $Y_1$ and $Y_2$ to have an exponential distribution it is necessary and sufficient to have that $Z_1$ and $Z_2$ to be independent. Am I interpreting what we have to prove incorrectly? We have to show that if $Y_1$ and $Y_2$ are independent and if $Z_1$ and $Z_2$ are independent then this implies that $X$ has a exponential distribution, and we also have to show that if $Y_1$ and $Y_2$ are independent and $X$ has an exponential distribution, then $Z_1$ and $Z_2$ are independent. – Some_Math_Nerd Oct 01 '20 at 23:16
  • $Y_1$ and $Y_2$ are never independent since $Y_1<Y_2$; it's a typo in the question. You have to show that $Z_1$ and $Z_2$ are independent if and only if $f$ is the pdf of $\Gamma(1,\beta)$. – StubbornAtom Oct 02 '20 at 03:53
  • Oh I did not realize it was a typo since the 7th and 8th edition of this text has the same exact problem – Some_Math_Nerd Oct 02 '20 at 05:36
  • It makes sense because when I try to show that if $X$ has a gamma pdf of $\alpha = 1$ and $\beta$ then $Y_1$ and $Y_2$ are independent, I was getting a contradiction. The article does not really help much because they assume $X$ and $Y$ are independent and that is what I replaced with for $Y_1$ and $Y_2$. – Some_Math_Nerd Oct 02 '20 at 05:38
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    Forget the article; it might be confusing you. Use the hint given in the exercise to prove the 'only if' part. The 'if part' is easy and has been answered here for example: https://math.stackexchange.com/q/2240822/321264. – StubbornAtom Oct 02 '20 at 07:51
  • I actually figured it out already, it was a typo. It was suppose to be $Z_1$ and $Z_2$ instead of $Y_1$ and $Y_2$. Thanks. – Some_Math_Nerd Oct 02 '20 at 08:42

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