Does the norm of the sum of $n$ 2d random vectors also scale as $\sqrt{n}$?

Question

Consider a 2d vector $\mathbf{V}$ which is obtained as the sum of $n$ i.i.d. standard normal random 2d vectors $\mathbf{v}_i$: $$ \mathbf{V} = \sum_{i=1}^n \mathbf{v}_i = \left( \sum_{i=1}^n x_i, \sum_{i=1}^n y_i \right)^T . $$

Now, I know that for a $n$-dimensional i.i.d. standard normal random vector $\mathbf{w} = (w_1,\ldots,w_n) \in \mathbb{R}^n$, the expected value of the norm $\mathbb{E}[\|\mathbf{w}\|]$ scales as $\sqrt{n}$, as stated, e.g., in this question or in this other one. But what about the expected value of the norm of $\mathbf{V}$, $\mathbb{E}[\|\mathbf{V}\|]=\mathbb{E}[\|\sum_{i=1}^n \mathbf{v}_i\|]$? Does it also scale as $\sqrt{n}$? If so, why? I am trying to wrap my mind around this, but I'm getting a bit lost...

Answer 1

If the $v_i=(x_i, y_i)$ are i.i.d. standard bivariate normal, then $\sum_{i=1}^n v_i$ has the same distribution as $nz$ where $z$ is standard bivariate normal.

Then, $E[\|nz\|] = n E[\|z\|] = cn$, so this scales as $n$ rather than $\sqrt{n}$.