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Consider some real-valued random variables $(X_n)$, $(Y_n)$, $X$ and $Y$, defined on the same probability space. What is a counterexample to the claim that $X_n\to X$ in distribution and $Y_n \to Y$ in distribution implies that $X_n + Y_n \to X + Y$ in distribution?

I know that Slutsky's theorem guarantees the implication when $Y = c$ holds, but not otherwise.

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  • $\begingroup$ Please add more details: are you dealing with probability spaces, general measure spaces, what are $X_n,Y_n,\ldots$? $\endgroup$ – Luiz Cordeiro Jun 16 '16 at 16:21
  • $\begingroup$ @LuizCordeiro Judging by the tags, I think we can assume we're working with probability spaces here. $\endgroup$ – Clarinetist Jun 16 '16 at 16:22
  • $\begingroup$ I guess convergence in distribution with independence works. $\endgroup$ – Clarinetist Jun 16 '16 at 16:29
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    $\begingroup$ @Clarinetist We shouldn't have to make assumptions about what the OP is asking; he should make things clearer in the question, as explained in meta.math.stackexchange.com/questions/9959/…, for example (of course, this is only my opinion, and everyone is feel to disagree). $\endgroup$ – Luiz Cordeiro Jun 16 '16 at 16:54
  • $\begingroup$ @LuizCordeiro : Doesn't the fact that Slutsky's theorem is cited answer your questions? $\qquad$ $\endgroup$ – Michael Hardy Jul 31 '16 at 6:30
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If the pair $(X_n,Y_n)$ converges in distribution to $(X,Y)$, then necessarily $X_n+Y_n$ will converge in distribution to $X+Y$ (by the continuous mapping theorem). So a counterexample would require us to specify $X$ and $Y$ and their joint distribution such that $(X_n,Y_n)$ does not converge in distribution to $(X,Y)$.

So take $X$ to be a nonconstant symmetric random variable, define $X_n:=X$, $Y_n:= X$, and $Y:=-X$. Then trivially $X_n$ converges in distribution to $X$, and $Y_n$ converges in distribution to $Y$ (since $X$ is symmetric). But $X_n+Y_n$ equals $2X$, while $X+Y=0$. Note that, as expected, $(X_n,Y_n)$ does not converge in distribution to $(X,Y)$, which is concentrated on the line $y=-x$; it converges in distribution to $(X,X)$, which is concentrated on the line $y=x$.

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Let $Z_k$ be iid with mean $0$ and variance $1$, and let $X_n = \frac{1}{\sqrt n} \sum_{k=1}^n Z_k$ and $Y_n = -X_n$. By the CLT, both $X_n$ and $Y_n$ converge in distribution to some $X$ which is standard normal. But $X_n + Y_n$ is always zero.

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I think the essence here is not convergence of sequences, but rather the difference between equality in distribution (denoted by $\sim$ in what follows) and almost sure equality (denoted by $=$ in what follows).

The following example, similar to what @grand_chat wrote down, should make it clear:

Let $Z \sim N(0, 1)$. Furthermore let $X = Y = A = Z$ and $B = -Z$.

Then since $Z \sim -Z$ we must have that $X \sim A$ and $Y \sim B$, but it does not hold that $X + Y \sim A + B$ since $X + Y \sim 2 \cdot N(0, 1)$ and $A + B = 0$.

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