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Let $X_1, \ldots, X_n$ be i.i.d. standard normal r.v. According to this answer, for the expected Euclidean norm of the random vector $X:=(X_1, \ldots, X_n)^T \in\mathbb R^n$ holds the following:

$$\frac{n}{\sqrt{n+1}}\le E(\|X\|)\le \sqrt{n}.$$

So $E(\|X\|)$ asymptotically equals to $\sqrt n$.

Is there some illuminating geometric explanation for that?

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It's the Pythagorean theorem, more or less. The quantity $\|X\|^2$ is the sum of the squares of the components, each of which is roughly 1 in magnitude. So $\|X\|^2$ is roughly $n$, and hence $\|X\|$ is roughly $\sqrt n$. One can use the Weak Law of Large numbers to conclude that the ratio $\|X\|/\sqrt n$ tends to 1 in probability, and the Central Limit theorem to get more precise estimates how much $\|X\|$ deviates from $1$ . And so on.

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  • $\begingroup$ Indeed! And is there any intuition on why $E(\|X\|)/\sqrt n$ is increasing? I.e. $E(\|X\|)/\sqrt n \to 1$ from below? $\endgroup$ – Leo Sep 9 '17 at 14:35
  • $\begingroup$ Not off the top of my head, without calculation. It should be obvious from some explicit integral formula or other, but... $\endgroup$ – kimchi lover Sep 9 '17 at 15:12
  • $\begingroup$ You could write $\|X\|^2/n = 1 + Z_n/\sqrt n$ where $Z_n\Rightarrow N(0,1)$, make the Taylor approximation $\|X\|/\sqrt n\approx 1 + Z_n/(2\sqrt{n}) - Z_n^2/(8n)+\dots$, and carry it under the integral to get $E\|X\|/\sqrt n\approx 1-1/(8n)$, but that would be a violate the law of never exchanging limiting operations unchecked. Or you could look hard at the Gamma functions appearing in the formula for the density function of the chi-squared distribution. $\endgroup$ – kimchi lover Sep 9 '17 at 17:29

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