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.

  • $\begingroup$ 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 $\endgroup$
    – dezdichado
    Sep 3, 2022 at 14:52

1 Answer 1


To provide a counterexample, the following Proposition(c.f. Y. S. Chow & H. Teicher, Probability Theory, 3rdEd, Springer Verlag, 1997, Prop. 8.4.3, p.299, or K. L. Chung, A Course in ProbabilityTheory, 3rdEd, Academic Press, 2003. Th.6.5.3, p.191.) is useful:

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.

  • $\begingroup$ Thank you very much. I really hoped for positive answer, so I didn't focus on looking for counterexamples. As expected, solution exploits multiplying by zero, which I couldn't cover. $\endgroup$
    – Esgeriath
    Sep 5, 2022 at 6:47

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