# Discrete correlation function (sample cross-covariance)

The continuous correlation function for the random variable $$A(t)$$ at a instant of time $$t$$ is 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 $$\tau < T$$. I'm trying to prove that by discritizing the variable $$\bar{t}$$, i.e., $$\bar{t}=i \Delta t$$ (integer $$i$$), 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.

• no one can help with what you wrote if you haven't even defined $d\bar{t}$ or what $A(\cdot)$ is Oct 21 '20 at 22:21
• Check your signs in the last equations. Whether your formula is correct depends on how you define the $A_i.$ Since the subscript $i-j$ in the final sum ranges from $1-j$ through $n-2j,$ which begins with values smaller than any appearing in your initial expression for $C_{AA}(j),$ your indexing looks suspicious.
• Problem is that one cannot choose an arbitrary $\tau$ and require convergence to an integral. In a construct of this type, a $\tau\to0$ would have to be imposed, which is inconsistent with the methods for $\tau$ selection in this question.