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  1. If $X_n\overset{p}{\rightarrow}X,Y_n\overset{p}{\rightarrow}Y$ then, show that $X_n+Y_n\overset{p}{\rightarrow}X+Y$.

I am not looking for the proof. My issue is, this seems contradictory to Slutsky's theorem, which is as follows.

  1. If $X_n\overset{p}{\rightarrow}X,Y_n\overset{p}{\rightarrow}c\in\mathbb{R}$ then, $X_n+Y_n\overset{p}{\rightarrow}X+c$. [Replacing all convergences in distribution by convergence in probability as mentioned on Wikipedia]. Also, for $Y_n\overset{p}{\rightarrow}c$, $c$ being a constant is a necessity otherwise, the result is not guaranteed.

However, if theorem 1 holds then this necessity is useless. Slutsky's theorem should be true regardless of $c$ being a constant or a random variable. Where am I wrong? Is it because the necessity is only valid if we only have $X_n\overset{d}{\rightarrow}X$?

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    $\begingroup$ $X_n \rightarrow X$ and $Y_n \rightarrow c$ implies $X_n+Y_n \rightarrow c$. This is the "converging together lemma", to give it a name. If $c$ is replaced by a non-degenerate random variable, the result does not remain true. $\endgroup$
    – Andrew
    Mar 12 at 12:28
  • $\begingroup$ @AndrewZhang so what you mean is theorem 1 is not valid? Couldyou share a counter example please? $\endgroup$
    – zaira
    Mar 12 at 12:29
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    $\begingroup$ Sorry, when I typed it, I used $\rightarrow$ instead of $\Rightarrow$, the latter is what I meant $\endgroup$
    – Andrew
    Mar 12 at 12:31

1 Answer 1

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Indeed a note on Wikipedia writes "The theorem remains valid if we replace all convergences in distribution with convergences in probability".

The same could be said if you replace all convergences with almost sure convergence, or $L^p$ convergence. As far as I am concerned I find this note completely useless, as it could be misunderstood for something deeper you think you would have missed.

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