# strong law of large numbers for V statistic

Recently, I encountered a problem regarding the a.s.-limit of $$\frac{1}{n^2} \sum_{k, \ell = 1}^n ||\boldsymbol X_k - \boldsymbol X_\ell||_2$$, where $$\boldsymbol X_i$$ are $$i.i.d.$$ sample following uniform distribution on $$[0, 1]^d$$. I am wondering how to prove/disprove its a.s.-convergence to $$\mathbb E[||\mathbf X_1 - \mathbf X_2||_2]$$.

The random variable mentioned above is a V-statistic with a symmetric kernel of degree 2, and it seems to me that my question is related to SLLN for V-statistic, but I couldn't find related literature on that (though there are plenty on the weak convergence of V-statistic).

Any ideas or suggestions would be really appreciated.

Note: I know U-statistic and V-statistic are asymptotically equivalent, and thus many weak convergence results of U-statistic can be concluded for the equivalent V-statistic. However, what I want to know here is a strong consistency result (a.s. convergence).

Update: This question has been solved. The solution is based on the suggestion from Ariel and this post. In my case, I can build equivalence between V-statistic and U-statistic, and then the a.s.-convergence result simply follows from the SLLN for U-statistic.

• The relationship between U and V statistics actually goes father then asymptotic equivalence. You can often represent a V-statistic as a U-statistic (see Section 3.2 of this paper). I think Van der Vaart (1998) also has an exercise on this. Perhaps, if this works for you then the standard Hoeffding SLLN for U-stats might work. Commented Jun 7, 2021 at 3:16
• This might also be helpful: math.stackexchange.com/questions/1908385/… Commented Jun 7, 2021 at 3:23
• @Ariel Thanks for pointing out this post! It turns out in my specific example, one can build exact equivalence between U-statistic and V-statistic, and the SLLN for V-statistic follows directly from Hoeffding's SLLN for U-statistic. Commented Jun 7, 2021 at 13:40

PROPOSITION 5.2.8. If $$\mathbb{E}|h|^{p}<\infty$$ with $$0, then $$\frac{1}{n^{m / p}} \sum_{1 \leq i_{1}<\cdots