I'm trying to do the following exercise:
Let $(x_i)_{1 \le i \le n}$ be a time series. Then the empirical auto-correlation function $\hat\rho: \{1,\ldots,n-1\} \to \mathbb R$ is defined by $$\hat\rho_n(k) = \frac{\sum_{i=1}^{n-k}(x_i -\bar x)(x_{i+k}-\bar x)}{\sum_{i=1}^{n} (x_i-\bar x)^2}$$
Fix $d \in \mathbb N_{>0}$ and $(a_0,\ldots,a_d) \in \mathbb R^{d+1}$. Now consider a particular time series in which $x_i = \sum_{j=0}^d a_j i^j$ for all $1 \le i \le n$. Prove that
For a fixed $k$, $\hat\rho_n(k) \longrightarrow 1$ as $n \longrightarrow \infty$.
I try to plug $$\bar x = \frac{1}{n} \sum_{j=0}^d a_j \sum_{i=1}^n i^j$$
in $(x_i -\bar x)(x_{i+k}-\bar x)$ and get $$\begin{aligned}(x_i -\bar x)(x_{i+k}-\bar x) &= \left (\sum_{j=0}^d a_j i^j -\frac{1}{n} \sum_{j=0}^d a_j \sum_{i=1}^n i^j \right) \left(\sum_{j=0}^d a_j (i+k)^j-\frac{1}{n} \sum_{j=0}^d a_j \sum_{i=1}^n i^j \right) \\ &= \left (\sum_{j=0}^d a_j \left ( i^j -\frac{1}{n} \sum_{i=1}^n i^j \right) \right) \left(\sum_{j=0}^d a_j \left ( (i+k)^j-\frac{1}{n} \sum_{i=1}^n i^j \right ) \right)\end{aligned}$$
Then I'm stuck at simplifying this expression and can not move on. Could you please elaborate on how to solve this problem?
The proof exploits the fact that the growth in $(x_i)$ is polynomial. To get $\hat\rho_n$ to look like $1$, the obvious step is to write $$(x_i-\bar x)(x_{i+k}-\bar x) = (x_i-\bar x)^2 + (x_i-\bar x)(x_{i+k}-x_i) $$ so that $$ \hat\rho_n=1-\frac{\sum_{i=n-k+1}^n(x_i-\bar x)^2}{\sum_1^n(x_i-\bar x)^2}+\frac{\sum_{i=1}^{n-k}(x_i-\bar x)(x_{i+k}-x_i)}{\sum_1^n(x_i-\bar x)^2}=:1-\varepsilon_n+\delta_n. $$ The aim is to show that both $\varepsilon_n$ and $\delta_n$ tend to zero. To do this, recall that $x_i= f(i)$ where $f$ is the polynomial function $$f(t):=\sum_{j=0}^da_j t^j.$$ We can assume wlog that the coefficient $a_d$ on the highest power $t^d$ is nonzero. It follows that, as $n\to\infty$: $$\begin{aligned} & x_i\sim a_d n^d\quad\text{if $i>n-k$}\\ &\sum_1^n x_i\sim \frac{a_d}{d+1}n^{d+1}\\ &\sum_1^n x_i^2\sim \frac{a_d^2}{2d+1}n^{2d+1}\\\end{aligned}$$ which in turn imply (as $n\to\infty$) $$\begin{aligned} &\bar x\sim\frac{a_d}{d+1}n^d\\ &x_i - \bar x\sim\frac {da_d}{d+1}n^d\qquad\text{if $i>n-k$}\\ &\sum_1^n(x_i-\bar x)^2=\sum_1^nx_i^2-n\bar x^2\sim \frac{a_d^2}{2d+1}n^{2d+1}-\left(\frac{a_d}{d+1}\right)^2n^{2d+1} \end{aligned} $$ and therefore $\varepsilon_n\to0$. To handle $\delta_n$, write $$ \delta_n^2\le \frac{\sum_{i=1}^{n-k}(x_i-\bar x)^2\sum_{i=1}^{n-k}(x_{i+k}-x_i)^2}{[\sum_1^n(x_i-\bar x)^2]^2}\le\frac{\sum_{i=1}^{n-k}(x_{i+k}-x_i)^2}{\sum_1^n(x_i-\bar x)^2} $$ where the first inequality is Cauchy-Schwarz. Bound the numerator of the RHS using the mean value theorem: $$|x_{i+k}-x_i|=|f(i+k)-f(i)|=|f'(i^*)k|, $$ where $i^*$ is between $i$ and $i+k$. Since $f'$ is a polynomial of degree $d-1$, the RHS of the above is at most $A+Bn^{d-1}$ for positive constants $A$ and $B$, so that $\sum_{i=1}^{n-k}(x_{i+k}-x_i)^2=O(n^{2d-1})$. Since $\sum_1^n(x_i-\bar x)^2=\Theta(n^{2d+1})$, conclude $\delta_n\to0$.