Understanding the Probability Distribution of Intensity Values

Question

From "Digital Image Processing 4th ed. - Gonzalez, Woods"

Let $p_r(r)$ and $p_s(s)$ denote the PDFs of intensity values $r$ and $s$ in two different images. A fundamental result from probability theory is that if $p_r(r)$ and $T(r)$ are known, and $T(r)$ is continuous and differentiable over the range of values of interest, then the PDF of the transformed (mapped) variable $s$ can be obtained as: $$(1)\ \ p_s(s)\triangleq p_r(r)\left|\frac{dr}{ds}\right|$$

I should specify that $T(r)$ indicates a single-pixel intensity transformation. Now, I understand what the two PDFs indicate, but I fail to understand:

1. The derivation of (1)
2. The utility of (1)

EDIT: I found that $(1)$ is basically the change of variable formula for probability density functions, which holds when $T(r)$ is monotonic.

Answer 1

Let $R$ be the random variable "intensity" with probability density function $f_R(r)$ ($=p_r$). That implies that $R$ also has a cumulative density function $F_R(r) = P(R \le r) = \int_{-\infty}^r f_R(u)\,\mathrm du$.

Let $S = T(R)$ be another random variable, the transformed version of $R$, with the transformation mapping $s=T(R)$ smooth and monotonous¹. We're looking for the pdf of $S$, $f_S(s)$ ($=p_s$).

We take the convenient detour through $f_S(s) = \frac{\mathrm d}{\mathrm ds}F_S(s)$:

\begin{align} \frac{\mathrm d}{\mathrm ds} F_S(s) &= \frac{\mathrm d}{\mathrm ds}P(S\le s)\\ &= \frac{\mathrm d}{\mathrm ds}P(S\le T(r))\\ &= \frac{\mathrm d}{\mathrm ds}P(T^{-1}(S)\le r)\\ &= \frac{\mathrm d}{\mathrm ds}F_R(T^{-1}(S)) & \text{chain rule of deriv.}\\ &= \frac{\mathrm d}{\mathrm d \left(T^{-1}(s)\right) }F_R(T^{-1}(S))\cdot \underbrace{\frac{\mathrm d}{\mathrm dS}T^{-1}(S)}_\text{deriv. of inv. function}\\ &=f_R(T^{-1}(S))\cdot \overbrace{\frac1{\frac{\mathrm d}{\mathrm dr}T(r)}}\\ &=f_R(R)\cdot \frac1{\frac{\mathrm d}{\mathrm dr}s}\\ &=f_R(R)\frac{\mathrm dr}{\mathrm ds} & \text{in original notation:}\\ &=p_r(r)\frac{\mathrm dr}{\mathrm ds} \end{align}

I feel like I've forgotten the formal step that introduces the absolute value, but in case of $T(r)$, I required local invertability, so that kind of goes with that. If someone has a good notation for that, please do not hesitate to edit my answer. It's kinda late!

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¹ that's an addition to the original problem, but for this way of derivation I need $T$ to be invertible. We can go a lot more measure-theoretic at this, but I don't really think that helps, so I'm going with this restriction, which I don't think is too harsh, as other mappings would lead to questionable results, just from a picture perspective.

Answer 2

Let $ R $ be a random variable and $ S = T \left( R \right) $ where $ T \left( \cdot \right) $ is a one to one continuous transformation on the domain of $ R $.

Since it is one to one transformation it must be strictly monotone. Let's test the 2 possible cases.

The Transformation $ T \left( \cdot \right) $ Is Strictly Increasing

Utilizing the function is one to one:

$$ {F}_{S} \left( s \right) = \mathbb{P} \left( S \leq s \right) = \mathbb{P} \left( T \left( R \right) \leq s \right) = \mathbb{P} \left( R \leq {T}^{-1} \left( s \right) \right) = {F}_{R} \left( {T}^{-1} \left( s \right) \right) $$

Using the Chain Rule to compute the density function:

$$ {f}_{S} \left( s \right) = {F}^{'}_{S} \left( s \right) = \frac{d}{d s} {F}_{R} \left( {T}^{-1} \left( s \right) \right) = {f}_{R} \left( {T}^{-1} \left( s \right) \right) \frac{d}{d s} {T}^{-1} \left( s \right) $$

The Transformation $ T \left( \cdot \right) $ Is Strictly Decreasing

Utilizing the function is one to one and decreasing:

$$ {F}_{S} \left( s \right) = \mathbb{P} \left( S \leq s \right) = \mathbb{P} \left( T \left( R \right) \leq s \right) = \mathbb{P} \left( R \leq {T}^{-1} \left( s \right) \right) = 1 - {F}_{R} \left( {T}^{-1} \left( s \right) \right) $$

Using the Chain Rule to compute the density function and the fact the inverse function also decreasing hence its derivative is also negative:

$$ {f}_{S} \left( s \right) = {F}^{'}_{S} \left( s \right) = -\frac{d}{d s} {F}_{R} \left( {T}^{-1} \left( s \right) \right) = {f}_{R} \left( {T}^{-1} \left( s \right) \right) \left(- \frac{d}{d s} {T}^{-1} \left( s \right) \right) $$

Summary

Since for the first case $ \frac{d}{d s} {T}^{-1} \left( s \right) $ is positive and for the second case $ - \frac{d}{d s} {T}^{-1} \left( s \right) $ is positive we can join them both into:

$$ {f}_{S} \left( s \right) = {F}^{'}_{S} \left( s \right) = -\frac{d}{d s} {F}_{R} \left( {T}^{-1} \left( s \right) \right) = {f}_{R} \left( {T}^{-1} \left( s \right) \right) \left| \frac{d}{d s} {T}^{-1} \left( s \right) \right| $$

Remark
I guess this is needed in the section before the Histogram Equalization where similar trick is used to derive the operator.