Retaining Possibilities Sincere: The Jacobian Adjustment

Introduction

buyer annoyance from wait instances. Calls arrive randomly, so wait time X follows an Exponential distribution—most waits are brief, a couple of are painfully lengthy.

Now I’d argue that annoyance isn’t linear: a 10-minute wait feels greater than twice as unhealthy as a 5-minute one. So that you determine to mannequin “annoyance models” as (Y = X²).

Easy, proper? Simply take the pdf of X, change x with (sqrt{y}), and also you’re executed.

You plot it. It appears affordable—peaked close to zero, lengthy tail.

However what when you truly computed the CDF? You’ll anticipate 1 proper?

The end result? 2.

import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import expon

# CDF of Exponential(1): F(x) = 1 - exp(-x) for x >= 0
def cdf_exp(x):
    return 1 - np.exp(-x)

# Mistaken (naive) pdf for Y = X²: simply substitute x = sqrt(y)
def wrong_pdf(y):
    return np.exp(-np.sqrt(y))  # This integrates to 2!

# Fast numerical verify of integral
from scipy.combine import quad
integral, err = quad(wrong_pdf, 0, np.inf)
print(f"Numerical integral ≈ {integral:.3f} (must be 1, but it surely's 2)")

# prints 2

Your new distribution claims each attainable final result is twice as seemingly correctly.

That’s unimaginable… but it surely occurred since you missed one small adjustment.

This “adjustment” is the Jacobian—a scaling issue that compensates for a way the transformation stretches or compresses the axis at completely different factors. Skip it, and your chances lie. Embody it, and every part provides up completely once more.

On this put up, we’ll construct the instinct, derive the maths step-by-step, see it seem naturally in histogram equalization, visualize the stretching/shrinking empirically, and show it with simulations.

The Instinct

To know why the Jacobian adjustment is important, let’s use a tangible analogy: consider a likelihood distribution as a hard and fast quantity of sand—precisely 1 pound—unfold alongside a quantity line, the place the peak of the sand pile at every level represents the likelihood density. The full sand at all times provides as much as 1, representing 100% likelihood.

Now, whenever you remodel the random variable (say, from X to Y = X²), it’s like grabbing that quantity line—a versatile rubber sheet—and warping it in accordance with the transformation. You’re not including or eradicating sand; you’re simply stretching or compressing completely different components of the sheet.

In areas the place the transformation compresses the sheet (a protracted stretch of the unique line will get squished right into a shorter phase on the brand new Y-axis), the identical quantity of sand now occupies much less horizontal area. To maintain the overall sand conserved, the pile should get taller—the density will increase. For instance, close to Y=0 within the squaring transformation, many small X values (from 0 to 1) get crammed right into a tiny Y interval (0 to 1), so the density shoots up dramatically.

Conversely, in areas the place the transformation stretches the sheet (a brief phase of the unique line will get pulled into an extended one on the Y-axis), the sand spreads out over extra space, making the pile shorter and flatter—the density decreases. For giant X (say, from 10 to 11), Y stretches from 100 to 121—a a lot wider interval—so the density thins on the market.

The important thing level: the overall sand stays precisely 1 lb, irrespective of the way you warp the sheet. With out accounting for this native stretching and shrinking, your new density could be inconsistent, like claiming you’ve gotten 2 lb of sand after the warp. The Jacobian is the mathematical issue that mechanically adjusts the peak all over the place to protect the overall quantity.

The Math

Let’s formalize the instinct with the instance of ( Y = g(X) = X^2 ), the place ( X ) has pdf ( f_X(x) = e^{-x} ) for ( x geq 0 ) (Exponential with fee 1).

Take into account a small interval round ( x ) with width ( Delta x ).
The likelihood in that interval is roughly ( f_X(x) Delta x ).

After transformation, this maps to an interval round ( y = x^2 ) with width
( Delta y approx left| g'(x) proper| Delta x = |2x| Delta x ).

To preserve likelihood:
$$ f_Y(y) Delta y approx f_X(x) Delta x, $$
so
$$ f_Y(y) approx frac{f_X(x)} g'(x) proper $$

Within the restrict as ( Delta x to 0 ), this turns into precise:
$$ f_Y(y) = f_X(x) left| frac{dx}{dy} proper|, $$
the place ( x = sqrt{y} ) (the inverse) and
( frac{dx}{dy} = frac{1}{2sqrt{y}} ).

Plugging in:
$$ f_Y(y) = e^{-sqrt{y}} cdot frac{1}{2sqrt{y}} quad textual content{for } y > 0. $$

With out the Jacobian time period ( frac{1}{2sqrt{y}} ), the naive
( f_Y(y) = e^{-sqrt{y}} )
integrates to 2:

Let ( u = sqrt{y} ), ( y = u^2 ), ( dy = 2u , du ):
$$ int_0^infty e^{-sqrt{y}} , dy $$

$$ = int_0^infty e^{-u}cdot 2u , du $$

$$= 2 int_0^infty u e^{-u} , du $$

$$ = 2 Gamma(2) = 2 cdot 1 = 2. $$

The Jacobian adjustment ensures $$ int_0^infty f_Y(y) , dy = 1. $$

This scaling issue ( left| frac{dx}{dy} proper| ) is exactly what compensates for the native stretching and shrinking of the axis.

The Basic Kind

Let Y = g(X), the place g is a strictly monotonic (rising or reducing) differentiable perform, and X has pdf ( f_X(x) ).

We would like the pdf ( f_Y(y) ) of Y.

Take into account a small interval round x with width ( Delta x ).
The likelihood in that interval is roughly ( f_X(x) Delta x ).

After transformation y = g(x), this interval maps to an interval round y with width
( Delta y approx left| g'(x) proper| Delta x ).

Going again to the equation we developed beforehand:
$$ f_Y(y) = f_X(x) left| frac{dx}{dy} proper|, $$
the place we use the inverse relation (x = h(y) = g^{-1}(y) ), and
( frac{dx}{dy} = h'(y) = frac{1}{g'(x)} ).

Thus the overall components is
$$ f_Y(y) = f_X(h(y)) left| h'(y) proper|. $$

Emperical Proof

Simulating the stretching and shrinking

The easiest way to “really feel” the stretching and shrinking is to zoom in on two areas individually: close to zero (the place compression occurs) and farther out (the place stretching dominates).

We’ll generate 4 plots:

1. Authentic X histogram, zoomed on small values (X < 1), with equal small intervals of width 0.1 — to indicate the supply of compression.

2. Corresponding Y = X² histogram, zoomed close to zero — exhibiting how these tiny X intervals get even tinier on Y (shrink).

3. Authentic X histogram for bigger values (X > 1), with equal intervals of width 1 — to indicate the supply of stretching.

4. Corresponding Y histogram for big values — exhibiting how these X intervals explode into enormous Y intervals (stretch).

import numpy as np
import matplotlib.pyplot as plt

# Generate massive pattern for clear visuals
n = 50000
x = np.random.exponential(scale=1, dimension=n)
y = x**2

fig = plt.determine(figsize=(16, 10))

def plot_histogram(ax, information, bins, density, shade, alpha, title, xlabel, ylabel):
    ax.hist(information, bins=bins, density=density, shade=shade, alpha=alpha)
    ax.set_title(title)
    ax.set_xlabel(xlabel)
    ax.set_ylabel(ylabel)

# Plot 1: X small values (compression supply)
ax1 = fig.add_subplot(2, 2, 1)
plot_histogram(ax1, x[x < 1], bins=50, density=True, shade='skyblue', alpha=0.7,
               title='Authentic X (zoomed X < 1)', xlabel='X', ylabel='Density')

# Equal-width intervals of 0.1 on small X
small_x_lines = np.arange(0, 1.01, 0.1)
for line in small_x_lines:
    ax1.axvline(line, shade='purple', linestyle='--', alpha=0.8)

# Plot 2: Y close to zero (exhibiting shrink/compression)
ax2 = fig.add_subplot(2, 2, 2)
plot_histogram(ax2, y[y < 1], bins=100, density=True, shade='lightcoral', alpha=0.7,
               title='Y = X² close to zero (compression seen)', xlabel='Y', ylabel='Density')

# Mapped small intervals on Y (very slim!)
small_y_lines = small_x_lines**2
for line in small_y_lines:
    ax2.axvline(line, shade='purple', linestyle='--', alpha=0.8)

# Plot 3: X bigger values (stretching supply)
ax3 = fig.add_subplot(2, 2, 3)
plot_histogram(ax3, x[(x > 1) & (x < 12)], bins=50, density=True, shade='skyblue', alpha=0.7,
               title='Authentic X (X > 1)', xlabel='X', ylabel='Density')

# Equal-width intervals of 1 on bigger X
large_x_starts = [1, 3, 5, 7, 9, 11]
large_x_lines = large_x_starts + [s + 1 for s in large_x_starts]
for line in large_x_lines:
    if line < 12:
        ax3.axvline(line, shade='purple', linestyle='--', alpha=0.8)

# Plot 4: Y massive values (exhibiting stretch)
ax4 = fig.add_subplot(2, 2, 4)
plot_histogram(ax4, y[(y > 1) & (y < 150)], bins=80, density=True, shade='lightgreen', alpha=0.7,
               title='Y = X² massive values (stretching seen)', xlabel='Y', ylabel='Density')

# Mapped massive intervals on Y (enormous gaps!)
large_y_lines = np.array(large_x_lines)**2
for line in large_y_lines:
    if line < 150:
        ax4.axvline(line, shade='purple', linestyle='--', alpha=0.8)

# Replace annotation kinds with modified font type
fig.textual content(0.5, 0.75, "X-axis shrinking → Density increasesnY-axis increased close to zero", 
         ha='middle', va='middle', fontsize=12, shade='black', 
         fontstyle='italic', bbox=dict(facecolor='#f7f7f7', edgecolor='none', alpha=0.9))

fig.textual content(0.5, 0.25, "X-axis stretching → Density decreasesnY-axis decrease for big values", 
         ha='middle', va='middle', fontsize=12, shade='black', 
         fontstyle='italic', bbox=dict(facecolor='#f7f7f7', edgecolor='none', alpha=0.9))

fig.set_dpi(250)
plt.tight_layout()
plt.present()

Simulating the Jacobian adjustment

To see the Jacobian adjustment in motion, let’s simulate information from the Exponential(1) distribution for X, compute Y = X², and plot the empirical histogram of Y in opposition to the theoretical pdf for rising pattern sizes n. As n grows, the histogram ought to converge to the proper adjusted pdf, not the naive one.

import numpy as np
import matplotlib.pyplot as plt

def correct_pdf(y):
    return np.exp(-np.sqrt(y)) / (2 * np.sqrt(y))

def naive_pdf(y):
    return np.exp(-np.sqrt(y))

# Pattern sizes to check
sample_sizes = [100, 1_000, 10_000]

fig, axs = plt.subplots(1, len(sample_sizes), figsize=(15, 5))

y_vals = np.linspace(0.01, 50, 1000)  # Vary for plotting theoretical pdfs

for i, n in enumerate(sample_sizes):
    # Pattern X ~ Exp(1)
    x = np.random.exponential(scale=1, dimension=n)
    y = x**2
    
    # Plot histogram (normalized to density)
    axs[i].hist(y, bins=50, vary=(0, 50), density=True, alpha=0.6, shade='skyblue', label='Empirical Histogram')
    
    # Plot theoretical pdfs
    axs[i].plot(y_vals, correct_pdf(y_vals), 'g-', label='Right PDF (with Jacobian)')
    axs[i].plot(y_vals, naive_pdf(y_vals), 'r--', label='Naive PDF (no Jacobian)')
    
    axs[i].set_title(f'n = {n}')
    axs[i].set_xlabel('Y = X²')
    axs[i].set_ylabel('Density')
    axs[i].legend()
    axs[i].set_ylim(0, 0.5)  # For constant viewing
    axs[i].grid(True)  # Add grid to every subplot

# Set the determine DPI to 250 for increased decision
fig.set_dpi(250)

plt.grid()
plt.tight_layout()
plt.present()

And the result’s what we anticipate.

Histogram Equalization: an actual world software

A traditional instance the place the Jacobian adjustment seems naturally is histogram equalization in picture processing.

We deal with pixel intensities X (usually in ([0, 255])) as samples from some distribution with empirical pdf based mostly on the picture histogram.

The purpose is to remodel them to new intensities Y in order that Y is roughly uniform on ([0, 255]) — this spreads out the values and improves distinction.

The transformation used is precisely the scaled cumulative distribution perform (CDF) of X:

$$ Y = 255 cdot F_X(X) $$

the place ( F_X(x) = int_{-infty}^x f_X(t) , dt ) (empirical CDF in observe).

Why does this work? It’s a direct software of the Likelihood Integral Rework (PIT):

If ( Y = F_X(X) ) and X is steady, then Y ~ Uniform([0,1]).

Scaling by 255 offers Uniform([0,255]).

Now see the Jacobian at work:

Let ( g(x) = L cdot F_X(x) ) (( L = 255 )).

The by-product ( g'(x) = L cdot f_X(x) ) (because the by-product of the CDF is the pdf).

Apply the change-of-variables components:

$$ f_Y(y) = f_X(x) / |g'(x)| = f_X(x) / (L f_X(x)) = 1/L $$

The ( f_X(x) ) cancels completely, leaving a relentless (uniform) density.

The Jacobian issue ( 1 / |g'(x)| ) mechanically flattens the distribution by compensating for areas the place the unique density was excessive or low.

In discrete photographs, rounding makes it approximate, however the precept is identical.

For a deeper dive into histogram equalization with examples, see my earlier put up: right here.

In Conclusion

The Jacobian adjustment is a kind of quiet items of arithmetic that feels pointless—till you skip it and immediately your chances don’t add as much as 1 anymore. Whether or not you’re squaring ready instances, modeling vitality from pace, or flattening picture histograms, the transformation modifications not simply the values however how likelihood is distributed throughout them. The issue ( left| frac{dx}{dy} proper| ) (or its multivariate cousin, the determinant) is the exact compensation that retains the overall likelihood conserved whereas accounting for native stretching and compression.

Subsequent time you remodel a random variable, bear in mind the sand on the rubber sheet: warp the axis all you need, however the whole sand should keep the identical. The Jacobian Adjustment is the rule that makes it occur.

Code

Hyperlink to Colab Pocket book

References and Additional Studying

A number of issues I discovered helpful.