Does $I(m,n) = I_{m,n}$?
I was looking at the formula for the mean-square error of two image and noticed they use $m$ and $n$ inside parentheses to denote rows and columns:
$MSE=\frac{\sum_{M, N}\left[I(m, n)-\hat{I}(m, n)\right]^2}{M * N}$
My question is would the following be equivalent?
$MSE=\frac{\sum_{M, N}\left[I_{m, n}-\hat{I}_{m, n}\right]^2}{M * N}$'