# Does the strong law of large numbers imply the convergence of moments of multivariate empirical distribution?

Suppose, $$X_1, \dots, X_n$$ are iid random vectors with some distribution $$P$$ on $$\mathbb{R}^d$$ and suppose $$E_P(|X_1^k|)$$ exists for all $$k \leq K$$ for some $$K > 1$$. Now, let $$\hat P_n$$ be the empirical distribution of $$X$$, constructed by putting weight $$1/n$$ on each observed $$X_i$$; let $$F_n$$ be its cdf.

If $$\dim(X) = 1$$, then by Glivenko-Cantelli, $$\Vert F_n - F\Vert_\infty \rightarrow 0$$ almost surely, and so, $$E_{\hat P_n}(X^k) \rightarrow E_P(X^k)$$, almost surely, for all $$k < K$$.

My question is, does this generalize to multiple dimensions? I know that the Glivenko-Cantelli result does not generalize directly without additional restrictions on $$P$$, but does almost sure convergence of moments hold for general $$P$$? If not, what kind of conditions are sufficient?

What about coordinate product moments? i.e. do we have $$E_{\hat P_n}(X_{J}^kX_{J}^{k'}) \rightarrow E_{P}(X_{J}^kX_{J}^{k'})$$ almost surely, where now $$X_{J}$$ represents the the sub-vector $$(X^{(j)})_{j \in J}$$ for some $$J \subset \{1, \dots, \dim(X)\}$$

EDIT: Does this simply follow from the Strong Law of Large numbers?

We should have $$E_{P_n}(X^k) = \sum_{i=1}^n X_i^k \xrightarrow{a.s.} E_{P}(X^k)~,$$ since $$\{X_i\}_{i=1}^n$$ is an iid sequence from $$P$$ and $$E(|X^{k}|)< \infty$$. Is this correct? I think this also directly generalizes to the coordinate product moments, no?

Changed title to reflect this.

• Coincidentally, when I'm free, I see your posts. Nov 22, 2020 at 1:51

Let $$\alpha$$ be any multi-index in $$\mathbb{N}^d$$ such that $$|\alpha|\le K$$
$$\mathbb{E}_{\hat{P}_N}(X^{\alpha}) = \dfrac{1}{N}\left( \sum_{n=1}^N X^{\alpha}_n \right)$$
Clearly, $$(X^{\alpha} ,n \ge 1)$$ is a sequence of integrable real random variables.
Hence, by the strong law oflarge number, we have: $$\dfrac{1}{N}\left( \sum_{n=1}^N X^{\alpha}_n \right) \longrightarrow \mathbb{E}_P(X^{\alpha}) \quad \text{a.e}$$
Hence, your conclusion. $$\square$$