# Finding the “expected” magnitude of the projection of a vector onto a random unit vector

In this 1-page article, the author claims that

the “expected” magnitude of the projection of a vector onto a random unit vector is proportional to its length [...] and it is nonzero.

However, I have emailed the author by asking how to calculate that expected value exactly, I got the response

[...]taking a surface integral over the n-dimensional sphere of the given quantity shown in the bracket.

So, if the proposed definition of the expected value function $E$, the integral has to be zero, because, we are evaluating a constant vector over S^n, and, for example, for $n=2$, we can employ the divergence theorem. Since the divergence of the constant vector is zero, the surface integral has to be zero. I mean even without the divergence theorem, for a constant vector $v$ and any unit vector $u$, $v \cdot u$ = - ( v \cdot (-u) ), so even intuitively I would expect this "expected" value to be zero.

So is there anything that I'm missing or just the author is somehow wrong in his claim ?

Edit:

I have sent my solution to the author, but got no response. • This is the expected value which is a fundamental concept in probability. It should appear in any textbook on probability. – angryavian May 17 '18 at 6:19
• @angryavian Yes, but in this case, the sample space, i.e our vector space, contains infinitely many element and that domain is not dense like the real line, so I'm having trouble for coming up with a meaningful definition in this case. – onurcanbektas May 17 '18 at 6:22
• I understand your question now. The distribution is the uniform distribution over the unit sphere (set of unit vectors). If you want something very concrete, take a standard Gaussian random vector $z$ and normalize it to get $z/\|z\|$; this normalized vector is uniformly distributed over the unit sphere. – angryavian May 17 '18 at 16:10