Consider two sequences of random variables, $\{X_n\}$ and $\{Y_n\}$. Let's assume $X_n\xrightarrow{D} X$, $Y_n \xrightarrow{D} Y$, and $\{X_n\}$ and $\{Y_n\}$ are independent of each other. It is well-known that under these conditions, $X_n + Y_n \xrightarrow{D} X+Y$ (one method of proof is to use the characteristic functions). But in real life, what matters is if $X_n + Y_m \xrightarrow{D} X+Y$ as $m,n\to \infty$ (what I mean by real life is when you are doing hypothesis testing and you have different sample sizes.) The easiest way I have been able to concoct a proof of the double sequnce result is to go through the first chapter of Billingsley's "Convergence of Probability Measures", and see if all the relevant results up to the statement about the convergence of product measures holds with double sequences. If I haven't missed any details, the proofs should carry over. But this is just too clunky. Is there an easier way to prove the double sequence analogue?

  • 1
    $\begingroup$ I think $X_n+Y_m = X_n + Y_n +(Y_m-Y_n) = X_n+Y_n + (Y-Y_n) + (Y_m-Y) \to X+Y+0+0$ $\endgroup$
    – SebastianZ
    Commented Jun 26, 2015 at 6:49
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    $\begingroup$ "It is well-known that under these conditions"... and under the condition that X and Y are independent. $\endgroup$
    – Did
    Commented Jun 26, 2015 at 7:37
  • $\begingroup$ @Did I agree, if $X$ and $Y$ aren't independent (say $X$ = $Y$), then funny stuff happens. Thanks for pointing it out. $\endgroup$
    – user765195
    Commented Jun 26, 2015 at 16:19
  • $\begingroup$ @SebastianZ, thank you for the elegant solution. I was stuck at the first equality, since $Y_n - Y_m$ doesn't converge to zero, but your second equality nails it and fixes the problem. $\endgroup$
    – user765195
    Commented Jun 26, 2015 at 16:21

1 Answer 1


As Did pointed out, we have to assume that $X$ and $Y$ are independent. Otherwise, consider a symmetric non-degenerated random variable $X$, $(X_n)$ and $(Y_n)$ independent identically distributed sequences where $X_1$ and $Y_1$ are distributed as $X$. Then $X_n\to X$ in distribution, $Y_n\to -X$ in distribution, but the sequence $(X_n+Y_n)_{n\geqslant 1}$ does not converge in distribution to $X-X=0$, since $X_1+Y_1$ is not degenerated.

In order to check the converge of $(X_n+Y_m)$ as $n,m$ go to infinity, we can use characteristic functions. That of $X_n+Y_m$ is $\mathbb E\left[e^{itX_n}\right]\cdot \mathbb E\left[e^{itY_m}\right]$ (by independence). This converges as $n,m$ go to infinite (say as $\min\{m,n\}\to \infty$) to $\mathbb E\left[e^{itX}\right]\cdot \mathbb E\left[e^{itY}\right]$, which is the characteristic function of $X+Y$, since $X$ and $Y$ are independent.


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