if $X,Y$ be a random independent variables if $X+Y$ and $Y$ has the same distribution, then $\mathbb{P}[X=0]=1$

Solution 1:

Use properties of characteristic functions.

NOTE. Variances, means MGFs etc cannot be used since they may not exist.

$Ee^{it(X+Y)}=Ee^{itX}Ee^{itY}$ by independence so

$Ee^{itY}=Ee^{itX}Ee^{itY}$. Fot $|t|$ sufficiently small we have $Ee^{itY} \neq 0$ so we get $Ee^{itX}=1$ for such $t$. This implies that $X=0$ a.s. (not $Y=0$ as stated)

[$E[1-\cos (tX)] =1-\Re \, (Ee^{itX})=0$ for $|t|<\delta$ implies that $1-\cos (tX)=0$ for $|t| <\delta$. So $X \in \{\frac {(2n+1) \pi} {2t}: n \in \mathbb Z\}$ a.s. for each $t$ with $0<|t|<\delta$. But $\{\frac {(2n+1) \pi} {2t}: n \in \mathbb Z\}$ and $\{\frac {(2n+1) \pi} {2s}: n \in \mathbb Z\}$ are disjoint if $\frac t s$ is irrational. Can you finish?].