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Suppose $U$ is uniformly distributed on $(0,1)$ then it is true that $X=\ln(1-U)\equiv \ln(U)$. This makes sense because $1-U\sim U$, and thus $X$ will have the same distribution. Now, suppose $c\in(0,1)$, then why can't I say that $$\ln(1-(1-c)U)\equiv\ln((1+c)U)$$ since $1-(1-c)U=1-U+cU$, and given that $1-U\sim U$, $1-U+cU=U+cU=(1+c)U$. Thus $1-(1-c)U\sim(1+c)U?$

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If $X \sim Y$ and $X' \sim Y'$ we cannot say $X+X' \sim Y+Y'$ without some independence assumptions. In this case $1-U$ and $cU$ are not independent (unless $c=0$) so the argumnet fails.

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  • $\begingroup$ Ahh, yes! I completely missed that. Thank you $\endgroup$ – DMH16 Oct 7 at 5:36

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