|WikiProject Statistics||(Rated B-class, Mid-importance)|
|WikiProject Mathematics||(Rated B-class, Mid-importance)|
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 184.108.40.206 (talk) 16:34, 2 January 2013 (UTC)
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.
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 220.127.116.11 (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 18.104.22.168 (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 22.214.171.124 (talk) 00:40, 12 April 2008 (UTC)