# Weak convergence of sum implies convergence of summands

Suppose $$X_n$$ and $$Y_n$$ are independent, $$X_n \overset d \to X$$ and $$X_n + Y_n \overset d \to Z$$ for some random variable $$Z$$.

Is it necessary, that $$Y_n$$ converges in distribution?

I tired using characteristic functions and Levy-Cramer theorem, but I'm missing one thing. Here is what I have: $$\forall t \in \mathbb R \quad \varphi_{X_n}(t) \to \varphi_X(t), \quad \varphi_{X_n}(t)\varphi_{Y_n}(t) \to \varphi_Z(t)$$ Since $$\varphi_{Y_n}$$ is bounded, it must converge for all $$t$$ such that $$\varphi_{X}(t) \neq 0$$. For those $$t$$ we would have $$\varphi_{Y_n}(t) \to \frac {\varphi_Z(t)}{\varphi_{X}(t)}$$ Which is continuous at $$0$$ and equal to $$1$$ there. If $$\varphi_{Y_n}$$ converged in every point, then $$Y_n$$ would have to converge in distribution.

Can I show that $$\varphi_{Y_n}$$ converges everywhere? Or in other way show that $$Y_n$$ converges in distribution? Counterexamples are welcome too.

• My probability is too rusty to recall anything immediately, but there is a nice book called Counterexamples in Probability which would probably have an answer to your question either direction. Amazon link Sep 3, 2022 at 14:52

Theorem(Pólya): A nonnegative, even function $$\psi$$ convex and decreasing on $$(0,\infty)$$ with $$\psi(0+)=\psi(0)=1$$ is a c.f.(characteristic function).
From Pólya's Theorem, the following $$\psi$$'s are c.f., for $$t\in\mathbb{R}$$ \begin{align*} \psi_1(t)&=(1-|t|)^+,\\ \psi_2(t)&=\Big(1-\frac{|t|}{2}\Big)^+,\\ \psi_3(t)&=\Big(1-\frac{|t|}{2}\Big)^+\vee \frac12. \end{align*} Now define the distibutions of $$X_n, Y_n$$ by c.f. as following: \begin{align*} \phi_{X_n}(t)&=\psi_1(t),\\ \phi_{Y_n}(t)&=\begin{cases} \psi_2(t), \quad & n=2k,\\ \psi_3(t), & n=2k-1. \end{cases} \end{align*} Then \begin{align*} \phi_{X_n+Y_n}(t)&=\phi_{X_n}(t)\phi_{Y_n}(t)\\ &=\psi_1(t)\psi_2(t). \end{align*} Hence $$X_n+Y_n\overset d\to Z$$, but the sequence of distributions of $$\{Y_{2k-1}, k\ge1\}$$ and sequence of distributions of $$\{Y_{2k}, k\ge 1\}$$ have different limit.