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.

  • $\begingroup$ Please add more details: are you dealing with probability spaces, general measure spaces, what are $X_n,Y_n,\ldots$? $\endgroup$ Jun 16 '16 at 16:21
  • $\begingroup$ @LuizCordeiro Judging by the tags, I think we can assume we're working with probability spaces here. $\endgroup$ Jun 16 '16 at 16:22
  • $\begingroup$ I guess convergence in distribution with independence works. $\endgroup$ Jun 16 '16 at 16:29
  • 3
    $\begingroup$ @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). $\endgroup$ Jun 16 '16 at 16:54
  • $\begingroup$ @LuizCordeiro : Doesn't the fact that Slutsky's theorem is cited answer your questions? $\qquad$ $\endgroup$ Jul 31 '16 at 6:30

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 $y=x$.


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.


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 $A + B = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.