Note I also asked this question on signal processing stackexchange If this question does not belong here as well I will delete it
I was reading a wiki page about signal averaging specifically in the section Signal power for sampled signals. At least at first glance the information seems to be quite wrong.
Passage from wikipedia follows:
Considering $n$vectors $V_i,\,i \in\{1,\ldots,n\}$ of signal samples of length $T$ $V_i = \left[s_{i,1}, \ldots, s_{i,T}\right],\quad s_{i,k} \in \mathbb K^T$,
the power $P_i$ of such a vector simply is
$P_i = \sum_{k=1}^T {s_{i,k}^2} = \left| V_i \right|^2$.
Again, averaging the $n$ vectors $V_{i},\,i=1,\ldots,n$, yields the following averaged vector
$V_\text{avg} = \frac 1n \sum_{k=1}^T\sum_{i=1}^n s_{i,k}= \frac 1n \sum_{i=1}^n \sum_{k=1}^T s_{i,k}$.
In the case where $V_n \equiv V_m \forall m,n \in \{ 1,\ldots, n\}$, we see that $V_\text{avg}$ reaches a maximum of
$V_\text{avg, identical signals} = P_i$.
In this case, the ratio of signal to noise also reaches a maximum,
$ \text{SNR}_\text{avg, identical signals} = \frac{V_\text{avg, identical signals}}{N_\text{avg}} = n \text{SNR}$.
This is the oversampling case, where the observed signal is correlated (because oversampling implies that the signal observations are strongly correlated).
Firstly, doesn't $$V_i = \left[s_{i,1}, \ldots, s_{i,T}\right],\quad s_{i,k} \in \mathbb K^T$$
Imply that $V_i$ is actually a vector of functions, $f:T\rightarrow K$ which is not what I think the intended meaning is. Shouldn't $V_i \in K^T$ and $s_{i,k} \in K$? Secondly, for $$V_\text{avg} = \frac 1n \sum_{k=1}^T\sum_{i=1}^n s_{i,k}= \frac 1n \sum_{i=1}^n \sum_{k=1}^T s_{i,k}$$ Isn't this just a scalar since they summed all the vectors and over the vectors? How is this the average vector? It may be that I don't understand something here but I feel like this section is full of mistakes and poorly explains whats going on. Am I correct that this is a mistake in the page or is there something I don't understand?