Histogram equalization - Contrast (gray value) distribution using transforming function

Question

I am looking for some guidance and explanation to be able to solve this exercise.

I understand that I have to come up with a function T(x) such that h(x) becomes H(x) and then with a U(x) which transforms back H(x) to h(x), but I am not sure on how to tackle this problem.

Exercise

Answer 1

I think this is more probability theory question. The trick is, if $x \sim F(x) $ ($F(x)$ is the cumulative distribution function or CDF in abbreviation form) then the transformation $ u = F(x)$ produces the uniform random variable $u$ with uniform distribution on interval $[0,1]$. you can see link below for proof:

Show Y has a uniform distribution if Y=F(X) where F(x)=P[X $\le$ x] is continuous in x.

Also to construct an arbitrary distribution $x \sim F(x)$ from a uniform random variable $u \sim U([0,1])$, it's enough to form the random variable $x = F^{-1}(u)$ (again $F(x)$ is the CDF of $x$). This one is the result of previous one if you change u with x. Therefore hre you need to apply the transformations to get the desired function:

So we have $h(x) = 6x^5 \Rightarrow F_x(x) = x^6 , x\in [0,1]$.

Then $y = x^6$ has the uniform distribution according to above lemmas.

To get a random variable with distribution $H(X) = 1.8X+0.1 \Rightarrow F_X(X) = 0.9X^2 + 0.1X$ you need to calculate the inverse of $F(X)$ which is $\frac{-0.1+\sqrt{0.81+4X}}{1.8}$ and then introduce the random variable $z = \frac{-0.1+\sqrt{0.81+4X}}{1.8}$ on the uniform random variable obtained from previous step. combining these two gives the transformation $t = \frac{-0.1+\sqrt{0.81+4x^6}}{1.8}$.

Therefore if $x$ is distributed as $6x^5$, then $t$, defined as $t = \frac{-0.1+\sqrt{0.81+4x^6}}{1.8}$ is distributed as $1.8t+0.1$.