# Expected Euclidean norm of a random vector

Let $$\mathbf{v} = (v_1,\ldots,v_n) \in \mathbb{R}^n$$ be a vector, whose coordinates are i.i.d random variables with zero mean and standard deviation $$\sigma$$.

Using Jensen's inequality, we obtain $${\mathbb{E}[\|\mathbf{v}\|]} \leq \sqrt{ \mathbb{E}[\|\mathbf{v}\|^2 } = \sqrt{ \sum_{i=1}^n \mathbb{E}[v_i^2] } = \sigma \sqrt{n}.$$

Question: Is this bound asymptotically sharp, i.e., does it hold that $$\lim_{n \to \infty} \frac{\mathbb{E}[\|\mathbf{v}\|]}{{\sqrt{n}}} = \sigma$$?

According to this answer, the strong law of large numbers "suggests" that this is the case. I am not sure, whether "suggests" is meant here in a heursitic sense, or whether it is supposed to mean "implies".

I am by no means an expert in probability theory, and therefore unfortunately fail to see, how the statement $$\lim_{n \to \infty} \frac{\mathbb{E}[\|\mathbf{v}\|]}{{\sqrt{n}}} = \sigma$$ would follow from the law of large numbers. As far as I understand it, the strong law of large numbers (together with the continuous mapping theorem) only implies that $$\sqrt{\frac{v_1^2+\ldots+v_n^2}{n}} \xrightarrow[]{a.s.} \sigma.$$ But as shown in this answer, almost sure convergence does not necessarily imply convergence of the mean.

You would need to use uniform integrability and the theorem that $$X_{n}\xrightarrow{L^{1}} X$$ if and only if $$\{X_{n}\}$$ is uniformly integrable and $$X_{n}\xrightarrow{P} X$$.

Let $$X_{n}=v_{n}^{2}$$ and $$S_{n}=\sum_{i=1}^{n}X_{i}$$ . Then, you already have $$\sqrt{\frac{S_{n}}{n}}\xrightarrow{a.s} \sigma$$ by the Strong Law (even weak law works here) and hence $$E[\sqrt{\frac{S_{n}}{n}}]\to \sigma$$ if and only if $$\sqrt{\frac{S_{n}}{n}}$$ is uniformly integrable.

If you look at my answer here , I have shown that $$\frac{S_{n}}{n}$$ is uniformly integrable. Hence you directly have that $$E[\frac{S_{n}}{n}]\to\sigma^{2}$$.

Then you also have that the sequence $$\sqrt\frac{S_{n}}{n}$$ is uniformly integrable.

Proof:-We use the following result .

Let $$\{X_{n}\}$$ be a sequence such that $$\{E[|X_{n}|^{p}]\}$$ is bounded for some $$p>1$$ , then $$\{X_{n}\}$$ is uniformly integrable. The proof of this is by the Markov inequality.

So here you have as $$E[|\sqrt{\frac{S_{n}}{n}}|^{2}]\to\sigma^{2}$$ and hence $$E[|\sqrt{\frac{S_{n}}{n}}|^{2}]$$ is bounded. And thus $$\sqrt{\frac{S_{n}}{n}}$$ is uniformly integrable.

Else you can use that if $$\{X_{n}\}$$ is uniformly integrable then , $$\{\sqrt{|X_{n}|}\}$$ is uniformly integrable. You'll need to use the Cauchy-Shwartz inequality . Or directly you can use the fact that the $$L^{p}$$ norm is weaker than the $$L^{q}$$ norm for finite measure spaces when $$p (the proof of which uses the Holder's Inequality) .

Thus you have the required result that $$E[\sqrt{\frac{S_{n}}{n}}]\to\sigma$$ .