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if $X_n$ converges to $X$ and $Y_n$ converges to $Y$ in distribution, what about $X_n + Y_n $ would that converge to $X+Y$ in distribution ? any ideas how i could prove or disprove this

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    It should be true if they are independent, see http://math.stackexchange.com/questions/591708/sum-of-two-independent-random-variables-converges-in-distribution – AlmostSureUser Nov 02 '15 at 16:50

3 Answers3

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Suppose that $X_n$ converges in distribution to $X$ where $X$ is a symmetric random variable, say $X \sim N(0,1)$. Then, trivially, $X_n$ also converges in distribution to $-X$ (since $X$ and $-X$ are identically distributed). However, $X_n + X_n$ does not converge in distribution to $X+(-X)=0$.

Shai Covo
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Convergence in distribution is a pretty weak concept.. Suppose you consider probability distributions on $[0,1]$. Let $X = Y = X_n$ for all $n$ have a density supported on $[0,1/2]$ alone, and let $Y_n$ be the same distribution except shifted to the right by $1/2$. Then all these random variables have the same distribution, so convergence in distribution is automatic. But also each $X_n + Y_n$ is the same, but different from $X + Y$, so you won't get convergence in distribution.

Zarrax
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  • Sorry but I do not get it: the measure obtained by shifting a distribution mu by a nonzero t cannot be mu itself. – Did Aug 10 '11 at 09:06
  • Sure but, in the example of your post, the random variables Y_n (almost surely in [1/2,1]) do not converge in distribution to the random variable Y (almost surely in [0,1/2]), although you sem to say they do. So, one wonders why X_n+Y_n should converge in distribution to X+Y anyway... – Did Aug 10 '11 at 14:39
  • Maybe I have the definitions wrong... isn't convergence in distribution just that for all $a$ $P(Y_n > a)$ converges to $P(Y > a)$ as $n$ goes to infinity?. You can make them both Gaussian to ensure $X_n + Y_n$ isn't the same as $X + Y$. – Zarrax Aug 10 '11 at 14:51
  • Precisely, P(Y_n>1/2)=1 for every n and P(Y>1/2)=0 hence the former does not converge to the latter. (Convergence in distribution is not exactly equivalent to what you write http://en.wikipedia.org/wiki/Convergence_of_random_variables#Definition_2 but let us ignore that.) About the second sentence of your comment: Gaussian random variables are clearly out of the scope of your post since no (nondegenerate) Gaussian is almost surely in [0,1]. – Did Aug 10 '11 at 15:18
  • You're not getting my definitions. The random variables can be unbounded, they're just defined on a measure space which I chose to be measurable subsets of $[0,1]$. At any rate, I'm done here. – Zarrax Aug 10 '11 at 15:25
  • Life is hard... :-) but the distribution of a random variable is a probability measure on its target space hence when one writes Suppose you consider probability distributions on [0,1] this can only mean that [0,1] is the target space of the random variables Xs and Ys. (By the way, as is often the case in probability, the source space (usually denoted Omega) is (nearly) irrelevant.) At this point, I simply do not see what your post means. – Did Aug 10 '11 at 15:39
  • In that case, the "source space" here is $[0,1]$, and the "target space" is ${\mathbb R}$. The definitons are just easier to describe with this source space. I promise this whole thing is quite easy. – Zarrax Aug 10 '11 at 15:47
  • If the source space is [0,1], how can the density of X be supported on [0,1/2] alone? Maybe you define X on half of its source space only? :-) Hmm, wait: are you changing P for P' when you replace X by Y? So which measure do you use for X+Y, P' or P? But then, what does something like P(Y>a) (which you wrote) mean? What a mishmash... The last sentence of your last comment is wonderfully ironical. (You know, I believe somehow my chakras are badly aligned these days, this is the second time I am led to (try to) explain these basic facts to somebody (who refuses to listen) on MSE. Oh well.) – Did Aug 10 '11 at 16:11
  • I don't really feel I have to put up with this crap. Here's another way to say exactly the same thing: $X_n = X = Y$ are one Gaussian random variable, and $Y_n$ is a second independent Gaussian random variable with the same distribution function. Figure it out. – Zarrax Aug 10 '11 at 16:17
  • Congratulations, this is an excellent example. Which has nothing in common with the setting of your post. – Did Aug 10 '11 at 16:26
  • Ok I kind of garbled what I meant: This should be accurate description of what I meant: Let $f(t)$ be some increasing nonnegative function, and let $X_n(t) = X(t) = Y(t) = f(t)$ for $0 \leq t \leq {1 \over 2}$, and zero for ${1 \over 2} \leq t \leq 1$, while $Y_n(t) = f(t - 1/2)$ for ${1 \over 2} \leq t \leq 1$ and $Y_n(t) = 0$ for $0 \leq t \leq {1 \over 2}$. Source space is $[0,1]$ and target space is ${\mathbb R}$ like before. I removed the garbled comment. – Zarrax Aug 10 '11 at 17:30
  • Also I should have said the random variable is supported on... rather than density. – Zarrax Aug 10 '11 at 17:37
  • Right. Let me precise and simplify this. First, one endows the set Omega=[0,1] with the Borel sigma-algebra and the Lebesgue measure P. Second, one can simply choose f(t)=1 for every t, hence X(t)=X_n(t)=Y(t) is 1 if t<1/2 and 0 if t>1/2 in [0,1]. Then X is what is called a Bernoulli random variable (this means that P(X=0 or 1)=1) and P(X=0)=P(X=1)=1/2. Let Y_n=1-X. Then Y_n is also Bernoulli hence (X_n) converges to X, (Y_n) converges to Y but P(X_n+Y_n=1)=1 and P(X+Y=0 or 2)=1 hence (X_n+Y_n) cannot converge to X+Y (all convergences in distribution). We happy. – Did Aug 10 '11 at 18:46
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The result works in the case that $Y_n \rightarrow c$ in distribution. This is because in this case $Y_n \rightarrow c$ in probability. Thus $P(|Y_n-c|>\epsilon)=P(|(X_n+Y_n)-(X_n+c)|>\epsilon) \rightarrow 0$. We deduce that $X_n+Y_n \rightarrow X_n+c$ in probability and hence also in distribution. But $X_n+c \rightarrow X+c$ in distribution (the proof is straightforward). So we have $X_n+Y_n \rightarrow X+c$ in distribution.

apatch
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