If a correlation matrix is a band matrix with value 1 on the band, is this equivalent to correlation matrix with all 1?

Question

Say I have a stationary sequence ${x_1,x_2,\cdots,x_n}$ and if two elements of the sequence are less than or $m$ apart from each other, we say they are correlated. This is called an m-dependent stationary sequence, whose correlation matrix would be a band matrix.

I am wondering what happens with the correlation between the neighboring points approach 1? Even if only adjacent elements are correlated with correlation 1, this would make all elements correlated with 1. (if $x_1$ is synchronized with $x_2$ and $x_2$ is synchronized with $x_3$, then $x_1$ and $x_3$ should be synchronized as well) This would mean that the correlation matrix is a square 1s matrix.

But band matrix (with 1s on the band) and 1s matrix are different (obviously). I am wondering how do we show the equivalency between these correlation matrices with mathematical formality?

Answer 1

Note that for square integrable random variables $X$ and $Y$, $$ \operatorname{Corr}(X,Y)=\pm 1 \quad\Leftrightarrow \quad X=a+bY $$ for some constants $a$ and $b$. Now, if $X_i$ and $X_{i+1}$ are perfectly correlated, then $$ X_{i+1}=a_i+b_i X_{i}. $$ Thus, we can write \begin{align} X_2&=a_1+b_1 X_1 \\ X_3&=a_2+b_2 X_2=a_2+b_2 (a_1+b_1X_1) \\ &=(a_2+b_2a_1)+b_2b_1X_1 \\ &\cdots, \end{align} or $$ X_i=a_i'+b_i' X_1, \quad i=2,\ldots, n $$ for some constants $a_i'$ and $b_i'\ne 0$. That is, $\operatorname{Corr}(X_i,X_j)=\pm 1$ for all $1\le i,j\le n$.