Talk:Slutsky's theorem

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 Field: Probability and statistics

old discussion…[edit]

The result in the article is not known as Slutsky's Theorem (that is a different result), but rather Slutsky's Lemma. The two results are cited often enough that the distinction should be made. — Preceding unsigned comment added by 98.223.197.174 (talk) 16:34, 2 January 2013 (UTC)

The claim is wrong for general X_n, Y_n. —Preceding unsigned comment added by 91.19.96.104 (talkcontribs)

Agreed. According to Fumio Hayashi's Econometrics textbook, Slutsky's Theorem says nothing about X_n and Y_n BOTH converging in distribution. Instead, if it is if X_n converges in distribution and Y_n converges in probability, then X_n + Y_n ... X_n*Y_n ... as stated already.

This is a very important distinction, but I'm not an expert. —Preceding unsigned comment added by 24.105.142.135 (talk) 16:30, 11 March 2008 (UTC)

Yes, I agree, this looks strange. One reference could be Bickel and Doksum, Mathematical statistics, theorem A.14.9, page 467. —Preceding unsigned comment added by 128.32.132.218 (talk) 19:29, 12 March 2008 (UTC)

- The theorem, as stated, only makes sense if one of the variables X or Y is constant. Otherwise the distributions of X_n+Y_n, X+Y, X_n.Y_n and X.Y are not properly defined ! To have a proper definition of these distributions, one would need the joint distributions of the (X_n,Y_n)'s and of (X,Y). Now if one of the limiting variables, say Y, is actually a constant, then Y_n converges to Y in distribution if and only if it converges to that constant in probability.

- In the third point (convergence of X_n/Y_n) it should be imposed that for n large enough, Y_n is almost surely non zero. Otherwise X_n/Y_n might not be defined on some non-negligible set of the probability space, even if Y \neq 0 a.s.

- To recap, the correct statement is : Let (Xn) and (Yn) be sequences of univariate random variables. If (Xn) converges in distribution to X and (Yn) converges in distribution to a constant a, then

   * (Xn + Yn) converges in distribution to X + a,
   * (XnYn) converges in distribution to a*X, and
   * (Xn / Yn) converges in distribution to X / Y if Y \neq 0 almost surely.

Note that the convergence in distribution of a sequence of random variables (Yn) to a constant a is equivalent to the same convergence in probability. —Preceding unsigned comment added by 81.57.2.37 (talk) 00:35, 12 April 2008 (UTC)

Sorry, the last point is: (Xn / Yn) converges in distribution to (X / a) if a \neq 0 and Yn \neq 0 almost surely for n large enough. —Preceding unsigned comment added by 81.57.2.37 (talk) 00:40, 12 April 2008 (UTC)

All these issues have been resolved, now the theorem’s statement is correct. ... stpasha » talk » 04:42, 13 September 2009 (UTC)