# show the variance here is bounded using the concentration of norm theorem

Let $$X\in{}\mathbb{R}^N$$, with independent sub-gaussian coordinates s.t. $$E[X_i^2]=1, E[X_i]=0$$.

W.T.S:

$$\text{Var}(\|X\|_2)\le{C'K^4}$$ with $$C'>0$$ and $$K:=\max_{1\le{i\le{N}}}{\|X_i\|_{\psi^2}}$$,

I don't know where to start, ive tried using various properties of sub-gaussian r.v. with no luck, any hints?

The concentration of the norm theorem:

Let $$X\in{}\mathbb{R}^N$$, with independent sub-gaussian coordinates s.t. $$E[X_i^2]=1$$. $$\|\|X\|_2-\sqrt{n}\|_{\Psi^2}\le{CK^2},\space{}C>0,\space{}K:=\max_{1\le{i\le{N}}}{\|X_i\|_{\psi^2}}$$

 



P.S. I have shown the fact that:

$$\sqrt{n}-CK^2\le{}\mathbb{E}[\|X\|_2]\le{}\sqrt{n}+CK^2$$

I essentially used the theorem, found a lower bound of the subgaussian norm of $$\|X\|_2-\sqrt{n}$$ with the Lp norm, set p to 1. and used Jensens inequality since $$f(x)=|x|$$ is a convex function.

I am using the following book and believe the question is similar to ex3.1.4: https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf

• The fact above with the assumptions provided is the concentration of the norm theorem, I will label it accordingly.
– kam
Commented May 11, 2020 at 16:37
• I will provide a sketch of the proof I have done, also I am using the text: 'High-Dimensional Probability: An Introduction with Applications in Data Science Book by Roman Vershynin'
– kam
Commented May 11, 2020 at 16:45
• Well actually, its from my lecture notes which are based on the book> I found it in the book as Exercise 3.1.4, but it's not quite the same. In the book we want to show that the variance is equal to $O(1)$.
– kam
Commented May 11, 2020 at 16:54
• Those are the notes I am using yes.
– kam
Commented May 11, 2020 at 17:00
• Presumably $\Psi$ and $||\cdot||_{\Psi^2}$ mean something. Commented May 11, 2020 at 17:07

Start with $$\text{Var}\|X\|_2 = \operatorname{E}(\|X\|_2 - \operatorname{E}\|X\|_2)^2 \leq \operatorname{E}(\|X\|_2 - \sqrt{n})^2$$ because the mean $$\text{E}\|X\|_2$$ minimizes the mean squared error. It is used for a similar question.

Further, by using the inequality for the $$L_1$$ norm $$|\|X\|_2 - \sqrt{n}| \leq CK^2$$ you obtained we have the upper bound as follows, $$\text{Var}\|X\|_2 \leq \text{E}(\|X\|_2 - \sqrt{n})^2 \leq C^2K^4 .$$

You can expand the definition of variance and use the mean linear property: \begin{aligned} \text{Var}(\|X\|_{2}) &= \mathbb{E}(\|X\|_{2} - \mathbb{E}\|X\|_{2} )^{2} \\ &= \mathbb{E}\|X\|_{2}^{2} - 2(\mathbb{E}\|X\|_{2})^{2} + (\mathbb{E}\|X\|_{2})^{2} \\ &= \mathbb{E}\|X\|_{2}^{2} - (\mathbb{E}\|X\|_{2})^{2} \\ \end{aligned} Since $$\mathbb{E}X_{i}^{2} = 1$$, $$\mathbb{E}\|X\|_{2}^{2} = n$$ and you can bound the second term using what you already have \begin{aligned} \text{Var}(\|X\|_{2}) &= \mathbb{E}\|X\|_{2}^{2} - (\mathbb{E}\|X\|_{2})^{2} \\ &\leq n - (\sqrt{n} + CK^{2})^{2} \\ &= -2\sqrt{n}CK^{2} + C^{2}K^{4} \\ &\leq C^{2}K^{4} \end{aligned}