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

 A: 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?].
A: 
Let $X,Y$ be a random independent variables if $X+Y$ and $Y$ has the same distribution,show that $\mathbb{P}[Y=0]=1$.

Sorry but there is a typo in the question
If $(X+Y)\stackrel{\text{d}}{=}Y$
Assuming the existence of MGF,  you get
$$\mathbb{E}[e^{Xt}]\mathbb{E}[e^{Yt}]=\mathbb{E}[e^{Yt}]$$
which implies that
$$\mathbb{E}[e^{Xt}]=1=e^{0t}$$
that is
$$\mathbb{P}(X=0)=1$$
not $\mathbb{P}(Y=0)=1$ as stated.

So, one of the two:

*

*If $(X+Y)\stackrel{\text{d}}{=}X$ then $\mathbb{P}(Y=0)=1$


*If $(X+Y)\stackrel{\text{d}}{=}Y$ then $\mathbb{P}(X=0)=1$

EDIT: for the sake of simplicity I used MGF but it may not exist. Thus it is a better choiche to use CF that always exist. The proof is exactly the same.
