Let $A$ be a $d\times d$ non-negative definite matrix. Let $\{{\bf{u}}_i\}_{i=1}^{n+1}$ be i.i.d. random vectors sampled uniformly at random from the unit sphere $\mathbb S^{d-1}\subset \mathbb R^d$.
Define $Q_n=\frac{1}{\sqrt n} \sum_{i=1}^n \textbf{u}_i^T A \textbf{u}_{i+1}.$
I want to calculate the expectations $\mathbb E[Q_n]$ and $\mathbb E[(Q_n)^2]$ in terms of $n,d, (a_{i,j})_{1\leq i,j \leq d}$. But I don't know how to proceed.
My attempt:
The pdf of every $\textbf{u}_i$ shall be $f_{\textbf{u}_i}(x_1,x_2,...,x_d)= \frac{1}{B}$ for all $(x_1,...,x_d) \in \mathbb S^{d-1}$ and equals zero otherwise, where $B$ is the area of $\mathbb S^{d-1}$. We may write $\textbf{u}_i$ as $(u_{i,1}, u_{i,2}, ..., u_{i,d})$ and $Q_n$ shall be a sum of terms of the form $u_{i,j} a_{j,k} u_{i+1,k}$ over $i,j,k$, where $1\leq i,j,k \leq d$.
We know $\{\textbf{u}_i\}_{i=1}^{n+1}$ is i.i.d. Can we conclude that $u_{i,j}$ and $u_{i+1,k}$ are independent, for all $i,j,k$ with $1\leq i,j,k \leq d$? (If it is true, can you give me a reference on this inference?) If they are really independent, then $\mathbb E[Q_n]=0$.
In the expression of $(Q_n)^2$, there will be terms like $u_{i,j} a_{j,k} u_{i+1,k} u_{i',j'} a_{j',k'} u_{i'+1,k'}$ for $1\leq i, i', j, j', k, k' \leq d$.
I don't know how to calculate $\mathbb E[u_{i,j} u_{i+1,k} u_{i',j'}u_{i'+1,k'}]$. Thanks for help.
Since the distribution of the $u_i$' s is invariant by rotation, without loss generality one may assume that $A=\mathrm{diag}(c_1,\ldots,c_d).$ Clearly $E(u_i)=0$ and therefore $E(Q_n)=\sum_{i=1}^nE(u_i)AE(u_{i+1})=0.$ Now $$Q_n^2=\sum_{i=1}^{n}E((u_iDu_{i+1})^2)+2\sum_{i<j}^{n}E((u_iDu_{i+1})\times u_jDu_{j+1}))=V+W$$ Clearly again $E(W)=0$ since $i<j$ implies that $u_i$ and $u_{j+1}$ are independent. Clearly also $E(V)=nE((u_1Du_{2})^2).$ For simplification write $u_1=(x_1,\ldots,x_d)$ and $u_2=(y_1,\ldots,d_d)$. We get using independence of $u_1$ and $u_2$ and the fact that $E(x_i)=E(y_j)=0$ $$E(V)=n \sum_ {k=1}^dE((x_kc_ky_k)^2)+2n \sum_ {k<m}E(x_kc_ky_k)\times (x_mc_my_m))=n \sum_ {k=1}^dE((x_kc_ky_k)^2).$$
We are left with the calculation of $$E((x_1c_1y_1)^2)=c_1^2(E(x_1^2))^2.$$ Since $x_1^2+\cdots+x_d^2=1$ we have $E(x_1^2)=1/d.$ Finally $$E(Q_n^2)=\frac{n}{d^2}(c_1^2+\cdots+c_d^2)=\frac{n}{d^2}\mathrm{trace}\, A^2.$$