Concentration of norm of linearly transformed normal random vector as dimension go to infinity

Question

Earlier asked on MSE, but didn't get an answer, so posting here:

Let $X=(X_1 \dots X_n) \in \mathbb{R}^n, X_i\sim N(0,1), iid.$ Let $B: \mathbb{R}^n \to \mathbb{R}^n $ be the diagonal linear map: $Bx_k:= x_k/ {k}, 1 \le k \le n.$ Then $||B||_F^2= \sum_{k=1}^{n}\frac{1}{k^2}$. Then is: $lim_{n \to \infty}|E||BX|| - ||B||_F |=0?$. How do we compute $E||BX||$ anyway? Note that, if $B$ were $I_n,$ the answer would be yes even for cases $X_i$ non Gaussian, c.f. this question on MO.

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Motivation for this question (not needed to answer the question): concentration

Note that: $E[||BX||^2]=||B||_F^2, ||.||_F$ denoting the Frobenius norm. For those familiar with concentration of measure OR Hanson-Wright inequality for concentration of quadratic forms, we could expect that $||BX||^2$ should be concentrated around $E[||BX||^2]=||B||_F^2.$ My question is: is the concentration asymptotically tight when dimension goes to infinity?

Following motivation from the fact: $lim_{n \to \infty}|E||X|| - \sqrt{n} |=0,$ I wonder: does $lim_{n \to \infty}|E||BX|| - ||B||_F |=0?$ Or if not, could we at least have: $\frac{||BX||}{||B||_F } \to _{p} 1$ in probability, as dimension $n \to \infty?$

I purposefully chose $B$ above so that the ratio of Frobenius norm to operator norm of $B$, i.e. $\frac{||B||_F}{||B||}$ does not go to $\infty.$ If it does go to infinity as $n \to \infty$, then we do have: $\frac{||BX||}{||B||_F } \to _{p} 1$, which follows from Hanson-Wright inequality. See the top/first equation from P.144 from this book.

Answer 1

The answer is no. Indeed, first of all, to make sense of the question, we need to deal with an infinite sequence of iid $N(0,1)$ random variables (r.v.'s) $X_1,X_2,\dots$. Next, for $n=1,2,\dots,\infty$, let $$Y_n:=\sqrt{\sum_1^n\frac{X_k^2}{k^2}}.$$ Your question can then be stated thus: Is it true that

$$EY_n\underset{n\to\infty}\longrightarrow\sqrt{EY_\infty^2}\,?\tag{1}$$

To answer this question, note first that, by the uniform integrability (see e.g. Corollary 12.8 and Proposition 12.9), $$EY_n\underset{n\to\infty}\longrightarrow EY_\infty.$$ So, the question becomes whether $\sqrt{EY_\infty^2}=EY_\infty$. But the latter equality may occur only if $P(Y_\infty=c)=1$ for some $c\in[0,\infty)$, which implies, in particular, that the r.v. $Y_\infty$ is discrete. In fact, however, the r.v. $Y_\infty$ is clearly absolutely continuous. Thus, the answer to question (1) is indeed no.