Let $X,Y\in\mathbb{N}_n^n$ be indexes, that is, $X=[0,2,1,...,3,n]$, namely, they are composed of enumerating from $0$ to $n$. $X,Y$ are independent and have a uniform distribution over all enumerations. Let $y_i$ be the $i$th component of $Y$.
Let $Z=\sum_{i=1}^nx_iy_i$ and let $k=\sum_{i=1}^ni^2$. I'd like to know the distribution of $W=\dfrac{Z}{k}$, or at least the mean and variance. Note that $0\leq W\leq1$ (there is a tighter lower bound).
As for the mean, this is what I did:
$$\mathbb{E}[W]=\mathbb{E}\left[\dfrac{Z}{k}\right]=\dfrac{1}{k}\mathbb{E}\left[\sum_{i=1}^nx_iy_i\right]=\dfrac{1}{k}\sum_{i=1}^n\mathbb{E}[x_iy_i]=\dfrac{n}{k}\mathbb{E}[x]\mathbb{E}[y]=\dfrac{n}{k}\dfrac{n}{2}\dfrac{n}{2}$$ $k=\dfrac{n(n+1)(2n+1)}{6}$. Then $\mathbb{E}[W]=\dfrac{6n^3}{8n^3+12n^8+n}$, for $n$ large enough the cube terms dominate, so $\mathbb{E}[W]=3/4$, which actually agrees with the experiments I ran.
Now, the variance is a trickier. Here is my attempt: $$\mathbb{E}[[W-\mu)^2]=\mathbb{E}[[\dfrac{Z}{k}-\mu)^2]=\dfrac{1}{k^2}\mathbb{E}[(Z-k\mu)^2]=\dfrac{1}{k^2}\mathbb{E}[Z^2-2k\mu Z+\mu^2]$$
My issues is calculating $\mathbb{E}[Z^2]$. This is what I have so far:
$$\mathbb{E}[Z^2]=\mathbb{E}\left[\left(\sum_{i=1}^nx_iy_i\right)^2\right]=\mathbb{E}\left[\sum_{i=1}^nx_i^2y_i^2+\sum_{i<j}x_iy_ix_jy_j\right]=\sum_{i=1}^n\mathbb{E}[x^2]\mathbb{E}[y^2]+\sum_{i<j}\mathbb{E}[x_ix_j]\mathbb{E}[y_iy_j]$$
Assuming what I did was correct, I am not sure what to do with $\mathbb{E}[x_jx_i]$. Can I assume they are independent since I am looking at different positions and say that is just $\mathbb{E}[x]\mathbb{E}[x]$, or instead $\mathbb{E}[x^2]$? or something else?
I'd appreciate if someone can confirm my analysis so far was correct and give me a hint on what to do about $\mathbb{E}[x_ix_j]$.
I attach the empirical distribution 
First, for $1\le i\le n+1$, $$ \mathsf{P}(x_i=a)=\frac{n!}{(n+1)!}=\frac{1}{n+1}, $$ and for $i\ne j$ and $a\ne b$, $$ \mathsf{P}(x_i=a,x_j=b)=\frac{(n-1)!}{(n+1)!}=\frac{1}{n(n+1)}. $$ Thus, $$ \mathsf{E}x_i^2=\sum_{k=0}^n\frac{k^2}{n+1}=\frac{n^2}{3}+\frac{n}{6}, $$ and for $i\ne j$, $$ \mathsf{E}x_ix_j=\sum_{k=0}^n\sum_{l\ne k}\frac{kl}{n(n+1)}=\frac{n^2}{4}-\frac{n}{12}-\frac{1}{6}. $$ Finally, since $x_ix_j\overset{d}{=}y_iy_j$, $$ \mathsf{E}Z^2=(n+1)\left(\frac{n^2}{3}+\frac{n}{6}\right)^2+n(n+1)\left(\frac{n^2}{4}-\frac{n}{12}-\frac{1}{6}\right)^2. $$