How Is the Bandwidth of an Image Reduced by an Octave in Gaussian Pyramid Procedure?
The model is simple.
Assume we have data I0 on the frequencies range of $\left[ 0, 1000 \right] $ [Hz].

We have 2 Black Boxes:


BlackBox 1 - LPF which always has a cut off at the middle of the spectrum.
BlackBox 2 - Decimation by 2.


What happens if we apply BlackBox 1 on the data?
We'll get an LPF version of it we'll call it L0 with data on the range $ \left[0, 500 \right] $ [Hz]..
We can also have H0 = I0 - L0 which is vasically all data on the range $ \left[500, 1000 \right] $ [Hz].

Now, be decimating I1 = decimate(L0) we have new data on the range $ \left[ 0, 500 \right] $ [Hz].
Applying the same tricks as above (Using the same black boxes which is what's nice about it) can yields the following series:


L# - Always contains the lower half frequency of the # Level.
H# - Always contains the higher half frequency of the # Level.


If you see the amount of data is reduced by factor of 2 each level (Octave).
Another nice property that you can reproduce the data using the lowest level LPF and the HPF of all other levels.