Average value of $x_1x_2+x_2x_3+x_3x_4+\cdots+x_{10}x_1$

Question

Average value of $S=x_1x_2+x_2x_3+x_3x_4+\cdots+x_{10}x_1$ where $x_1x_2,x_3...$ are all possible permutations of $1,2,3,...10$

My try:

$S$ is sum of the products taken two at a time of $x_1$,$x_2$,...$x_{10}$

so we can write $$2S=(x_1+x_2+x_3+\cdots x_{10})^2-(x_1^2+x_2^2+\cdots x_{10}^2)$$

But since for all possible permutations of $1,2,3..10$, the sum of them and sum of squares of them remains fixed.

Hence average of $S$ is $S$ itself.

So we get

$$2S=(1+2+3+..10)^2-(1^2+2^2+3^2+..10^2)$$

with that we get $$S=1320$$

EDIT: the actual sum $S$(ten terms) is:

$$S=x_1x_2+x_2x_3+\cdots +x_9x_{10}+x_{10}x_1$$

Now how to proceed?

Answer 1

In order to compute $$\sum_{\sigma\in S_{10}}\sigma(1)\sigma(2)+\ldots+\sigma(10)\sigma(1)$$ let us focus on the sum $$ \sum_{\sigma\in S_{10}} \sigma(1)\sigma(2) $$ first. This sum contains all the products of the form $ab$, with $a$ and $b$ being distinct elements of $\{1,\ldots,10\}$, each one appearing $8!$ times. It follows that $$ \sum_{\sigma\in S_{10}} \sigma(1)\sigma(2) = 8!\sum_{a\neq b}ab = 8!\left[\left(\sum_{a=1}^{10}a\right)^2-\sum_{a=1}^{10}a^2\right]=8!\cdot2640. $$ Then you can check that $\sum_{\sigma\in S_{10}}\sigma(2)\sigma(3)$ is exactly the same number and draw your conclusion.
If my computations are correct, the average value of your quadratic form is $\frac{880}{3}$.

Answer 2

When I compute your final equation I get $1320$, not $1410$. In your original, does $S$ just have $10$ terms, each of the form $x_kx_{k+1}$ (and the last cyclic one) or did you assume all $45$ terms of the form $x_mx_n$ with $n \neq m$? The title and first line suggest only $10$, but your later work has $45$.

Answer 3

Since$$\Bbb Ex_1x_2=\frac{1}{2\binom{10}{2}}\left(\left(\sum_{i=1}^{10}i\right)^2-\sum_ii^2\right)=\frac{55^2-\frac1610\cdot11\cdot21}{90}=\frac{88}{3}$$and similarly with the other terms, and means are additive even for correlated terms, $\Bbb ES=\frac{880}{3}$. (If we replace $10$ with $n$, you can show this generalizes to $\Bbb ES=\frac{n(n+1)(3n+2)}{12}$.)

Answer 4

Let $x_{11} = x_1$ and treat $x_1,\ldots,x_{10}$ as random variables. Taking average over all permutation of $x_k$ is equivalent to taking expectation values over them. $$\begin{align}\verb/E/\left[\sum_{k=1}^{10} x_k x_{k+1}\right] &= \sum_{k=1}^{10} \verb/E/[ x_k x_{k+1} ] = 10 \verb/E/[ x_1 x_2]\\ &= 10 \verb/E/_{x_1}\big[ x_1 E[x_2|x_1]\big]\\ &= \frac{10}{10}\sum_{n=1}^{10} n\left(\frac{55-n}{9}\right) = \frac19 \sum_{n=1}^{10} n(55 -n)\\ &= \frac19 \left[55^2 - \frac{10(10+1)(2\cdot 10+1)}{6}\right]\\ &= \frac{880}{3} \end{align} $$