+\qquad \( \sum_{i=0}^{L-1}{p(x_{i})} = 1 \)
+
+Trivially, the function $p(x_{i})$ can be thought of as counting the recurrences of the value $x_{i}$ in the $bin_i$.
+The histogram is similar to a bar graph (not the same: the histogram uses continuous data, the bar graph uses discrete data) and on the abscissa it shows $x_{i}$ values and on the ordinate $y_{i}$. Each x in a range value is treated as independent of neighboring values so it is considered as an isolated unit called an "accumulator" or "bin." It is on the bins that we count the occurrence of the $x_{i}$ value ($max_i$), which gives us precisely the value $p(x_{i})$. The Histogram is called a bar graph because a value of x is actually an interval between $x_{i}$ and the next value $(x_{i+\varepsilon})$. Because x has continuous values computed in floating point and normalized interval $0 - 1.0$ is used, there is no solution of continuity between one bin and another and the boundaries are decided a priori, usually based on bit depth color. The bin concept is fundamental because it is the basis on which we can do mathematical calculations. In fact, the area of the bin is the frequency (count or $max_i$) in which that value occurs; the width of the bin is the different values within the range we consider. With 8 bits of depth color we have 256 bins; then we collect the values of x from the initial value 0 up to and including 1; then from 1 up to and including 2; and so on up to the last bin, which ranges from 254 to and including 255. It is clear, then, that the continuous luma values are bounded in a range and made to become discrete values on which it is easier to perform calculations. The width of a bin is given by the formula:
+
+$width = a_i = x_{max_i} - x_{min_i}$
+
+Having established the depth color, the bin width is always the same (a) for every $bin_i$:
+
+$a_i = a = \dfrac{range}{\# bins}$
+
+For a depth color of 8 bits we have (normalized range $0 - 1.0$):
+
+$a_{8bit} = \dfrac{(1.0-0)}{256} = 1/256$
+
+For a depth color of 10 bits or more we have:
+
+$a_{10bit} = \dfrac{(1.0-0)}{65536} = 1/65536$
+
+Wider bins have a higher count (because they gather more $x_{i}$). Narrower bins have a lower $max_i$ (because they contain less $x_{i}$; neighboring values ($x_{i+\varepsilon}$) are distributed in neighboring bins).
+To recap: in \CGG{} histogram is a bunch of \textit{bins} (accumulators) that count the number of times a particular pixel channel intensity (luma, $x_{i}$) occurs in an image. The plugin scans all the pixels in the frame, counting the frequencies in each given bin ($max_i$). Knowing the width of the bins (a) then it is easy to get the height of $bin_i$ (y). In fact, the bins are rectangles, and you can apply the area formula from which to derive the height:
+
+$A_i = max_i = Base \times High = a \times y_i$
+
+Hence:
+
+$y_i = \dfrac{max_i}{a}$
+
+Dim bins are on the left, bright bins on the right.
+You can have discordance of results, looking in the scopes, either by switching from Histogram to Histogram Bezier or after a conversion between color spaces (with associated change in depth color). The number of bins used depends on the color model bit depth: