# Theorem contradicting the necessity of Slutsky's theorem

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 $$X_n\overset{d}{\rightarrow}X$$?

• $X_n \rightarrow X$ 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. Mar 12 at 12:28
• @AndrewZhang so what you mean is theorem 1 is not valid? Couldyou share a counter example please? Mar 12 at 12:29
• Sorry, when I typed it, I used $\rightarrow$ instead of $\Rightarrow$, the latter is what I meant Mar 12 at 12:31

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.