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Suppose we have two real random variables $X$ and $Y$ which are independent and $X+Y$ has the same distribution as $X$. Does this imply that distribution of $Y$ is degenerate at 0?

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Compute $E[X+Y]$ and $E[(X+Y)^2]$ and see where that gets you. –  GEdgar Jun 22 '13 at 14:07
I know that if X,Y \in L^{2}, then it's easy, but what about the general case? –  Josef Jun 22 '13 at 16:49
Maybe characteristic functions can be used to prove this. Characteristic function of Y would be constantly 1. –  Josef Jun 24 '13 at 7:23
Just an aside: Note that we can't dispense with the independence assumption in general. For example, let $X$ have a symmetric distribution about 0 and take $Y = -2 X$. Then, $X \stackrel{d}{=} X+Y$ but $Y \neq 0$ a.s. –  cardinal Jun 26 '13 at 1:32
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Let $\phi(t) = E(e^{itX})$ and $\psi(t) = E(e^{itY})$ be the characteristic functions of $X$ and $Y$. By hypothesis, we have $$\forall t \in \Bbb R,\qquad \phi(t)(1-\psi(t))=0$$

The function $\phi$ being continuous with $\phi(0)=1$, we can find an open neighborhood $U$ of $0$ where $\phi$ does not vanish. Therefore $\psi(t)=1$ for every $t \in U$, which implies $tY \in 2\pi\Bbb Z$ a.s for all $t \in U$.

By density of $\Bbb R\setminus\Bbb Q$ in $\Bbb R$, we can find $t_1,t_2 \in U\setminus\{0\}$ such that $t_1/t_2 \notin \Bbb Q$. Then, with probability one, we have $$ Y \in (2\pi/t_1)\Bbb Z \cap (2\pi/t_2)\Bbb Z = \{0\}. $$

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Could you explain why we can conclude that $tY \in 2 \pi Z$? –  robinson Jun 26 '13 at 9:54
If $E(e^{itY})=1$, then $E(1-\cos(tY))=0$ hence $\cos(tY) = 1$ a.s. because $1-\cos \geq 0$. –  Siméon Jun 26 '13 at 11:42
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If you allow the so-called improper random variables then we can assume the probability distribution X with all the mass at infinity. Now, a distribution of Y can be any distribution not assuming infinite values with positive probability, i.e. any proper distribution. Thus, in the above case, a distribution of Y does not have to be degenerate at 0.

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