If $X \rightarrow_d aY$ and $W \rightarrow_d Y$ where $a$ is a constant in $\mathbb{R}$.
Is
$$X + W \rightarrow_d (a + 1)Y$$
true due to the fact that the limiting random variable is the "same"?
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3
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3$\begingroup$ Suppose $X$ and $W$ are independent and identically distributed with finite nonzero variance $\sigma^2.$ Set $X_i=X$ and $W_i=W,$ so that trivially $X_i\to X = (1)Y$ and $W_i\to W=Y$ where $Y$ has the same distribution as $X$ and $W.$ (Thus $a=1.$) Notice that $\operatorname{Var}(X+W)=2\sigma^2$ but $\operatorname{Var}((a+1)Y) = (a+1)^2\operatorname{Var}(Y)=2^2\sigma^2.$ Observe that $2\ne 2^2$ and draw your conclusions. $\endgroup$– whuber ♦Oct 20, 2022 at 12:35
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1$\begingroup$ Do I read your comment correctly that you have two r.v.s $W,X$ having the same distribution, viz. that of $Y$, where (possibly) the OP means that $Y$ is the same r.v. in either case? If there is a difference, would it matter? $\endgroup$– Christoph HanckOct 20, 2022 at 13:02
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1$\begingroup$ @ChristophHanck OP has specified convergence in distribution (AFAIK) with the subscripted d under the arrow, so I think there is only one interpretation. $\endgroup$– seanv507Oct 20, 2022 at 15:08
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