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I'm trying to show that: $$ (X_n,Y_n)\to^p(X,Y)\iff X_n\to^pX,Y_n\to^p Y $$ where $\to^p$ means convergence in probability ($P(||X_n-X||>\varepsilon)\to 0,\forall\varepsilon>0$).

I managed to show $(\Rightarrow)$, but I don't know how to show $(\Leftarrow)$.

Question: How to prove that $$\{||(X_n,Y_n)-(X,Y)||>\varepsilon\}\subseteq\{||X_n-X||>\frac{\varepsilon}{2}\}\cup\{||Y_n-Y||>\frac{\varepsilon}{2}\} $$

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  • $\begingroup$ @GuihermeSalome I need to show the same thing but I don't know how to show the $(\Rightarrow)$ direction. It seems like it should be obvious that if $(X_{n},Y_{n})\to^{P} (X,Y)$ that we should have $X_{n} \to^{P} X$ and $Y_{n} \to^{P} Y$, but I'm sure it's not that simple. How did you do it? $\endgroup$ – NotThatGrumpyAnymore Dec 19 '17 at 3:24
  • $\begingroup$ @GuihermeSalome also, if you have to prove that $\{||(X_n,Y_n)-(X,Y)||>\varepsilon\}\subseteq\{||X_n-X||>\frac{\varepsilon}{2}\}\cup\{||Y_n-Y||>\frac{\varepsilon}{2}\}$, do you also have to prove that $\{||X_n-X||>\frac{\varepsilon}{2}\}\cup\{||Y_n-Y||>\frac{\varepsilon}{2}\} \subseteq \{||(X_n,Y_n)-(X,Y)||>\varepsilon\}$? $\endgroup$ – NotThatGrumpyAnymore Dec 19 '17 at 3:44
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If this is all you're trying to prove let's rewrite some things and see if that helps. I assume we are working with Euclidean distance here.

Suppose $\|(X_n,Y_n) - (X,Y)\| > \epsilon$ and $\|X_n - X\| \leq \epsilon/2$. We want to show that $\|Y_n - Y\| > \epsilon/2$, then by symmetry we're done.

So we're assuming \begin{eqnarray} (X_n - X)^2 + (Y_n - Y)^2 &>& \epsilon^2, \\ (X_n - X)^2 &\leq& \epsilon^2/4. \end{eqnarray} And we are trying to show that

$$ (Y_n - Y)^2 > \epsilon^2/4. $$

Should I let you take it over from here?

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  • $\begingroup$ if you would't mind completing this proof, I would be very much appreciative! How does this help us? $\endgroup$ – NotThatGrumpyAnymore Dec 19 '17 at 3:25
  • $\begingroup$ also, if you could answer the questions I asked the OP above, because I doubt he cares to revisit a question he asked 2 years ago, I would be forever grateful to you! :) $\endgroup$ – NotThatGrumpyAnymore Dec 19 '17 at 3:46
  • $\begingroup$ @Jeff hey there, I might be wrong butwhat you are asking might be Slutky’s theorem. $\endgroup$ – Guilherme Salomé Dec 19 '17 at 14:44
  • $\begingroup$ @GuilhemeSalome it's related to Slutsky's Theorem, but it's not exactly. $\endgroup$ – NotThatGrumpyAnymore Dec 19 '17 at 16:09
  • $\begingroup$ @Jeff I found what theorem it is (we call it the "magic" theorem here haha). Shoot me an email at guilhermesalome at gmail.com and I'll send you a pdf with it. $\endgroup$ – Guilherme Salomé Dec 19 '17 at 21:47

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