Let $Z$ be a real-valued random variable, and suppose that $Z = X_1 + X_2$ where $X_1$ and $X_2$ are i.i.d. random variables. Suppose further that $Z = Y_1 + Y_2$ where $Y_1$ and $Y_2$ i.i.d. random variables. Does it follow that $Y_1$ and $X_1$ are identically distributed?
I've looked into indecomposable distributions and infinitely divisible distributions, but could not find a result/example immediately answering the above question.
Best Answer
The answer is no in general.
In the language of characteristic functions, you have $\phi_Z(t)=\phi^2_X(t)=\phi_Y^2(t)$. Taking the square root implies $\phi_Y(t)=\pm\phi_X(t)$ for every $t$, in particular with $\phi_X(t)=\phi_Y(t)$ for $t$ in a neighborhood of 0 (since $\phi(0)=1$ for any characteristic function and by uniform continuity).
Thus, this is true if you require that $\phi_X$ and $\phi_Y$ are analytic, as they would necessarily agree on an interval, and therefore everywhere.
Otherwise it is not true, meaning that $|\phi_X(t)|=|\phi_Y(t)|$, such that there are $t$ where $\phi_X(t)\neq \phi_Y(t)$. A classic example of this is as follows:
Let $P(X=x)=\frac{2}{\pi^2(2k-1)^2}$ whenever $x=\pm (2k-1)\pi$ for $k=1,2\cdots$ , $P(X=0)=1/2$, and $P(X=x)=0$ otherwise.
Let $P(Y=y)=\frac{4}{\pi^2(2k-1)^2}$ whenever $y=\pm (2k-1)\pi/2$ and $k=1,2,\cdots$ and $P(Y=y)=0$ otherwise.
Both these give rise to "triangle-wave" characteristic functions that are periodic outside the interval specified:
$$\phi_X(t)=1-|t|, t\in[-1,1], \mbox{ periodic otherwise}$$ $$\phi_Y(t)=1-|t|, t\in[-2,2], \mbox{ periodic otherwise},$$
which look like: