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Sep 21 at 2:20 comment added Mingzhou Liu Hi, I am wondering why $\phi_\mu\neq 0$ implies that there exiets a $f$ such that $\phi_\mu=e^f$?
Jan 4, 2021 at 5:04 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Dec 5, 2020 at 4:59 answer added AndreyF timeline score: 1
Nov 15, 2020 at 21:48 comment added Mateusz Kwaśnicki This assumes "good will" of the reader, I believe. Compare with the following statement: If $\sqrt{x} = 1$, then $x = 1$. This is usually assumed to be true, but you could argue that if $x < 0$, then what is meant by $\sqrt{x}$? Here, in a similar way, you should read the definition of $\psi$ as "Suppose that the characteristic function of $\mu$ is non-zero, and its continuous logarithm is denoted by $\psi$."
Nov 15, 2020 at 11:44 history asked 0xbadf00d CC BY-SA 4.0