If $X+Y$ has the same distribution as $X$, is $Y$ almost surely zero?

Solution 1:

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 $$ Y \in (2\pi/t_1)\Bbb Z \cap (2\pi/t_2)\Bbb Z = \{0\}. $$

Solution 2:

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.