Assume $\sqrt{n}(X_n-X)\overset{d}{\to} N(0,\sigma^2)$ and $Y_n\overset{p}{\to} c$, where $X$ is a random variable and $c$ is a positive constant. Does it hold that $\sqrt{n}(Y_nX_n-cX)\overset{d}{\to} N(0,c^2\sigma^2)$?
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StubbornAtom
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John
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3No. Let $c=1, X=1,Y_n=1+1/n^{1/4}$. – Michael Aug 29 '20 at 07:29
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1It would work if $\sqrt{n}|X(Y_n-c)|\rightarrow 0$ in probability. – Michael Aug 29 '20 at 07:41
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@Michael I'm having trouble coming to terms with this. Why would Michael's counterexample work? Doesn't Slutsky's theorem state that the above would hold? – user577215664 Jul 11 '21 at 18:38
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@MtGlasser : You can try to apply Slutsky's theorem (somehow?) and then see what happens and what goes wrong. – Michael Jul 11 '21 at 23:23
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@mj33 Thanks Michael.... – user577215664 Jul 11 '21 at 23:48
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@Michael You might want to look at the linked question https://math.stackexchange.com/questions/4196191 where I may have said things you could say better – Henry Jul 12 '21 at 01:37