0
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ no one can help with what you wrote if you haven't even defined $d\bar{t}$ or what $A(\cdot)$ is $\endgroup$
    – develarist
    Oct 21 '20 at 22:21
  • $\begingroup$ Thanks, I've just addressed your comments. $\endgroup$
    – sined
    Oct 21 '20 at 22:33
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Oct 23 '20 at 20:42
  • 1
    $\begingroup$ 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. $\endgroup$
    – Carl
    Oct 29 '20 at 10:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.