I am wondering what is the $\mathbb{E}[(x\cdot v)^2]$ where $x$ is a random unit vector in $\mathbb{R}^n$ and $v$ is a given unit vector in $\mathbb{R}^n$. By $(x\cdot v)$ I mean the dot product between $x$ and $v$.
I read somewhere that $\mathbb{E}[(x \cdot v)^2] = \frac{1}{n}$. But I can not prove it. Does anyone have an idea on this?
Judging from the result, it appears the context implicitly supposes the distribution of $x$ is invariant under orthogonal transformations: I would call this a spherically-symmetric distribution.
(There are plenty of spherically-symmetric distributionns. Starting with any distribution $F$ in $\mathbb{R}^n$, define $\tilde F$ to be the values of $F$ averaged over the action of the orthogonal group $O(n)$ of rotations and reflections about the origin. The average exists because $O(n)$ is compact and acts continuously. It is immediate that $\tilde F$ is invariant under $O(n)$. In particular, a Normal distribution of mean $(0,0,\ldots,0)$, diagonal variance matrix, and equal variances is spherical.)
It is a (simple and geometrically obvious) algebraic result that any unit vector $v$ can be extended to an orthonormal frame $(v=v_1, v_2, \ldots, v_n)$. Because the distribution is spherically symmetric and any element $v_i$ can be rotated into any other element $v_j$, the coordinates $x\cdot v_i$ all have the same distribution. Let $\mu_2$ be the common expected value of all the $(x\cdot v_i)^2$.
Since $x$ is assumed to be a unit vector,
$$1 = 1^2 = x\cdot x = \sum_i (x\cdot v_i)^2.$$
Take expectations of both sides and use the linearity of expectation to compute
$$1 = \mathbb{E}(1) = \mathbb{E}(x\cdot x) = \mathbb{E}\left( \sum_{i=1}^n (x\cdot v_i)^2\right) = \sum_{i=1}^n \mathbb{E}((x\cdot v_i)^2) = \sum_{i=1}^n \mu_2 = n\mu_2,$$
implying $1/n = \mu_2 = \mathbb E ((x\cdot v)^2)$.
