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I am trying to show the following: $X_n \Rightarrow X$ and $Y_n \Rightarrow c$, c a constant, implies $X_nY_n \Rightarrow Xc$.

Using the fact that $X_n \Rightarrow X$ and $Y_n \Rightarrow c$, c a constant, implies $X_n + Y_n \Rightarrow X+ c$ (which I know how to prove) I found in another post (Proving Slutsky's theorem) a way to prove what I want to prove :

"We have $$X_nY_n=X_n(Y_n-c)+cX_n;$$ defining $Z_n:=cX_n$ and $Z'_n:=X_n(Y_n-c)$, we reach the wanted conclusion provided that we manage to show that $X_n(Y_n-c)\to 0$ in probability. But for a fixed $\varepsilon$, and each $R$ $$\mathbb P\{| X_n(Y_n-c)|\gt \varepsilon\}\leqslant\mathbb P\{|X_n|\gt R\}+\mathbb P\{|Y_n-c|\gt \varepsilon/R \}.$$ Choosing $R$ as a limiting point of the distribution function of $|X|$, we obtain from the convergence of $Y_n$ to $c$ in probability that $$\limsup_{n\to +\infty} \mathbb P\left\{| X_n(Y_n-c)|\gt \varepsilon\right\}\leqslant \mathbb P\{|X|\gt R\}.$$ Since $R$ can be chosen arbitrarily large, we are done. "

I agree with the proof except the last step, to conclude that $P\{|X|\gt R\}$ goes to zero when R is arbitrary large we need the hypothesis that X is finite almost surely but I have no such hypothesis. Is there a way to modify the proof to work without any such hypothesis on X ?

P.S: I want to prove my statement without using the fact that $(X_n,Y_n) \Rightarrow (X,c)$.

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  • $\begingroup$ If $X$ is a random variable then it is a function $\Omega\to\mathbb R$ so by definition it is finite a.s. For that a hypothesis is not needed. $\endgroup$
    – drhab
    Apr 12, 2018 at 8:01
  • $\begingroup$ Unfortunately in my course we work in the extended real line, Random variable can take value at infinity. $\endgroup$
    – DimSum
    Apr 12, 2018 at 9:48
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    $\begingroup$ Is Slutsky valid in that situation? What e.g. if $X_n=n$ and $Y=\frac1n$? then $X_n\stackrel{d}{\to} X$ where $P(X=\infty)=1$ and $Y_n\stackrel{p}{\to} 0$. This with $X_nY_n\stackrel{p}{\to} 1$. But can we state that $0\cdot\infty=1$? $\endgroup$
    – drhab
    Apr 12, 2018 at 9:54
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    $\begingroup$ No we can't, so if I follow your reasoning, we HAVE TO make the assumption that X is finite almost surely to obtain $X_n \Rightarrow X$ and $Y_n \Rightarrow c$, c a constant, implies $X_nY_n \Rightarrow Xc$. $\endgroup$
    – DimSum
    Apr 12, 2018 at 10:20
  • $\begingroup$ Yes. Working in $[-\infty,\infty]$ Slutsky is only valid under extra conditions. Another condition might be that $|c|\neq\infty$. $\endgroup$
    – drhab
    Apr 12, 2018 at 10:27

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