# Slutsky's lemma; same limit

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"?

• 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.
– whuber
Oct 20, 2022 at 12:35
• 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? Oct 20, 2022 at 13:02
• @ChristophHanck OP has specified convergence in distribution (AFAIK) with the subscripted d under the arrow, so I think there is only one interpretation. Oct 20, 2022 at 15:08