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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...

Tropilio
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  • For clarity, you should at least state some properties of the distribution (i.i.d. coordinates with zero mean and common variance?). Also, why do you specify that they are 2D vectors when you are interested in its behavior as the dimension $n$ changes? – angryavian Jul 12 '23 at 15:38
  • @angryavian Yes, thanks, I have updated the question. The vectors are all 2d, it's only the number of vectors $n$ being added together that changes. – Tropilio Jul 12 '23 at 15:51

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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 $\sqrt{n}$.

angryavian
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  • Thank you! I am a bit lost however... How would it be in 1d? If I have $v_1, v_2, \dots, v_n$ i.i.d. standard normal random numbers, what would be the expected value of the norm of the sum $E[ | \sum_{i=1}^n v_i | ]$? Would it scale as $n$ or $\sqrt{n}$? – Tropilio Jul 13 '23 at 08:00
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    @Tropilio It would also scale as $n$. By a similar argument, $\sum_{i=1}^n v_i$ has distribution $nz$ where $z$ is standard normal. So $E[|nz|] = n E[|z|] = (\sqrt{2/\pi})n$. – angryavian Jul 13 '23 at 15:46
  • All right. Thank you very much! For completeness, I add that if instead the $v_i$ are uniformly distributed in the interval $[0,1]$, then their sum follows the Irwin-Hall distribution, with expected value $n/2$. – Tropilio Jul 14 '23 at 10:16