*Earlier asked on [MSE](https://math.stackexchange.com/questions/3646822/concentration-of-norm-of-linearly-transformed-normal-random-vector-as-dimension), 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](https://mathoverflow.net/questions/325862/expectation-of-the-norm-of-a-random-vector?rq=1). 




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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](https://en.wikipedia.org/wiki/Concentration_of_measure) OR [Hanson-Wright inequality for concentration of quadratic forms](https://arxiv.org/abs/1306.2872), 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](https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html#).



Concentration of norm of linearly transformed normal random vector as dimension go to infinity
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)](https://web.ma.utexas.edu/users/gordanz/notes/uniform_integrability.pdf), 
$$EY\_n\underset{n\to\infty}\longrightarrow EY\_\infty.$$
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. 


