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 $$\{||(X_n,Y_n)-(X,Y)||>\varepsilon\}\subseteq\{||X_n-X||>\frac{\varepsilon}{2}\}\cup\{||Y_n-Y||>\frac{\varepsilon}{2}\} $$

  • $\begingroup$ @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? $\endgroup$ Dec 19, 2017 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 $\{||X_n-X||>\frac{\varepsilon}{2}\}\cup\{||Y_n-Y||>\frac{\varepsilon}{2}\} \subseteq \{||(X_n,Y_n)-(X,Y)||>\varepsilon\}$? $\endgroup$ Dec 19, 2017 at 3:44

1 Answer 1


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?

  • $\begingroup$ if you would't mind completing this proof, I would be very much appreciative! How does this help us? $\endgroup$ Dec 19, 2017 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! :) $\endgroup$ Dec 19, 2017 at 3:46
  • $\begingroup$ @Jeff hey there, I might be wrong butwhat you are asking might be Slutky’s theorem. $\endgroup$ Dec 19, 2017 at 14:44
  • $\begingroup$ @GuilhemeSalome it's related to Slutsky's Theorem, but it's not exactly. $\endgroup$ Dec 19, 2017 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. $\endgroup$ Dec 19, 2017 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.