Autocorrelation function between received and training signal used for channel estimation
I think it's just a trick which makes $R(t)=0, \mathrm{when}\quad t&lt;0$. 

First, $-N+1$ is just a time offset and not affect the essence of function. 

$r_{CE}(t)$ is bounded to $[0, 2N+T_{CH}-2]$, and $a_N(t)$ to $[0, N-1]$. In the form you gave, $R(t)$ is bounded to $[-N+1, 2N+T_{CH}-2]$. May be the author want the non-zero value of $R(t)$ also starts at 0, so he just move it right with $N-1$ length, which causes the $-N+1$ term in the equation.

--------- Why $R(t)$ is bounded as that -------------

This is just a explanation to give some intuition, not a formal proof. Consider two function $f(t)$ and $g(t)$, which $f(t)=0, t \notin [a, b]$ and $g(t)=0, t\notin [c, d]$. Now we want to calculate the sum of products, i.e., $S=\sum_t f(t)g(t)$, $[a,b]$ and $[c,d]$ must be overlapped, right? Thus we get $a \le d $ and $c\le b$.

Get back to the $R(t)=\sum _d r_{CE}(t+d)a^*(d)$(note here the argument is $d$), $a^*(d)$ is bounded to $[0, N-1]$ and $r_{CE}(t+d)$ to $[-t, L-1-t]$, in which $L=2N+T_{CH}-1$ is the length of $r_{CE}(d)$. Here we get $a=0, b=N-1, c=-t, d=L-1-t$. Using $a \le d $ and $c \le b$, we get $-(N-1) \le t \le L-1$.

As to why $r_{CE}$ is bounded as that, you can make a same process to understand it.