# Histogram equalization

I seems to be stuck in one of my derivation regarding histogram equalization. The other StackExchange answers didn't help me go through it. I would like someone to check if my derivation/set up is correct as well as help me finish.

So the problem is that we have an image with intensities values in $[0,1]$. We set the probability density function of the image as $p:[0,1] \rightarrow \mathbb{R}$. We set the distribution function as $P:\sigma([0,1]) \rightarrow [0,1]$. We also set a transition function from old intensity to new ones as $g:[0,1] \rightarrow [0,1]$. Now from probability theory we have : \begin{align} P(x < a) = \int_0^a p(t)dt = F(a) \end{align} With $F$ being the cumulative distribution function. Now let's assume $y = g(x)$ where y is the transformed intensity and assume g is monotone increasing. We have : \begin{align} F_2(a) = P(y < a) = P(g(x) < a) = P(x < g^{-1}(a)) = F(g^{-1}(a)) \end{align} Where $F_2$ is the cumulative distribution function of the transformed intensities. Now let's take derivatives : \begin{align} F_2'(a) &= F(g^{-1}(a))' \\ &= F'(g^{-1}(a)).(g^{-1})'(a)\\ &= p(g^{-1}(a)).(g^{-1})'(a) \end{align}

We want the density of the transformed pixel to be constant so $F_2'(a) = (\int_0^a dt)' = 1$ So that : \begin{align} 1 = p(g^{-1}(a)).(g^{-1})'(a) \end{align}

Now in all solutions I've found they say to take $g^{-1}(a) = \int_0^a p(t)dt$. But when plugging in to see if everything checks out I'm left with: \begin{align} 1 = p\big( \int_0^a p(t)dt\big).p(a) \end{align}

What am I missing here ?

## 1 Answer

Ok the problem seems to come from my sloppy notation. The probability function $P$ can't be used to model both histogram. To be able to do that I need to be more precise and use random variable. Here is the solution I came out with :

• Let $\Omega = [0,N-1]$ be the probability space equipped with the usual sigma algebra. $N$ is the number of pixels in the image.
• Let X and Y be Random variable such that $X:\Omega \rightarrow [0,1]$ and $Y:\Omega \rightarrow [0,1]$. X and Y represent the assignment of a pixels intensity (continuous case ). X is going to model the current image, and Y is going to represent the image with the flat histogram.
• Let $g:[0,1] \rightarrow [0,1]$ represent the mapping function from old intensities to new ones. With these definitions we have $Y(\omega) = g(X(\omega))$. Now let's calculate the cumulative distribution function of Y :

\begin{align} P_Y(y<b) &= PY^{-1}(y<b) \\ &= P(\{\omega \in \Omega | Y(\omega) < b\})\\ &= P(\{\omega \in \Omega | g(X(\omega)) < b\})\\ &=P(\{\omega \in \Omega | X(\omega) < g^{-1}(b)\}) \text{ g is monotone increasing}\\ &= PX^{-1}(y<g^{-1}(b)) \\ &=\int_0^{g^{-1}(b)} f(t)dt \end{align}

On the other hand since Y has a flat density we have $P_Y(y<b) = b$ so that : \begin{align} b &= \int_0^{g^{-1}(b)} f(t)dt\\ g(b) &= \int_0^{b} f(t)dt\\ \end{align}

Does that sounds right ?