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How can be the expected length of projection of vector of length $l$ is $l/\sqrt d$ where d is dimension of underlying vector space. I am using $l_2^2$ norm

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  • $\begingroup$ For d=2 the expected length of the projection of a unit vector on any axis is 2/pi, not 1/sqrt(2). Please explain why you suggest 1/sqrt(d). $\endgroup$ – Did Apr 13 '14 at 9:57
  • $\begingroup$ i am talking about $l_2^2$ norm. I edited my post please check it $\endgroup$ – thetatheta Apr 15 '14 at 13:13
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Let $x_i$ denote the $i$th coordinate of a vector uniformly distributed on the sphere of radius $\ell$, then $E(x_i^2)$ does not depend on $i$ and $\sum\limits_{i=1}^dx_i^2=\ell^2$ with full probability hence $E(x_i^2)=\ell^2/d$ for every $i$, that is, $\|x_i\|_2=E(x_i^2)^{1/2}=\ell/\sqrt{d}$. Note that $\|x_i\|_2$ is not the expected length of the projection on the $i$th axis (this would be $\|x_i\|_1=E(|x_i|)$).

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