If I start with a continuous correlation function given by
\begin{equation} C_{AA}(\tau) =\frac{1}{T} \int_{0}^T d\bar{t} A(\bar{t})A(\bar{t}+\tau) \end{equation} with $\tau < T$. How can I prove that by discritizing the variable $\bar{t}$, i.e., $\bar{t}=i \Delta t$, with $T=N \Delta t$, I obtain
\begin{equation} C_{AA}(j)=\frac{1}{N} \sum_{i=1}^{N-j} A_i A_{i+j} \label{eq1} \end{equation}
I started by making the following substitution of variables $t\rightarrow\bar{t}+\tau$, $dt=d\bar{t}$, yielding
\begin{equation} C_{AA}\left(\tau\right) = \frac{1}{T}\int_{0}^{T-\tau}dtA\left(t-\tau\right)A\left(t\right) \end{equation}
Using now $t\rightarrow t_{i}=i\Delta t$ and $\tau=j\Delta t$ (with $T=\Delta t\times N$) we obtain \begin{equation} C_{AA}\left(j\right) =\frac{1}{\Delta t\times N}\sum_{i=1}^{N-j}\Delta tA_{i\Delta t-j\Delta t}A_{i\Delta t} =\frac{1}{N}\sum_{i=1}^{N-j}A_i A_{i-j}\end{equation}
However, this formula desagrees slightly from the equation I'm supposed to obtain.
You didn't find the right boundaries of integration, after the sub. the boundaries are $\tau$ and $T+\tau$.