Uniform and Gaussian distributions — Mathematical Foundations

Compare flat and Gaussian densities

PDF comparison · solid selected, dashed alternative

x · density has inverse-x units · drawn tails are truncated, total Gaussian mass is analytically one

CDF · accumulated area

F(x) = area to the left · approaches zero and one in the tails

Gaussian mean 0, standard deviation 1. Selected closed interval [-1, 1]. A PDF height is density, with inverse-value units. Interval probability is area: approximately 0.682689. A single continuous point has probability zero. Numerical CDF absolute error budget 1e-9; interval budget 2e-9. The finite drawing does not prove normalization.

Selected interval area: 0.682689
CDF at lower bound: 0.158655
CDF at upper bound: 0.841345
Density at Gaussian mean: 0.398942
Any exact point probability: 0
Standardized lower bound: −1
Standardized upper bound: 1

= exact model quantity · ≈ numerical companion or observed estimate

Probability representation

Uniform start

from −10 to 10, up to 2 decimal places

Uniform end

from −10 to 10, up to 2 decimal places

Gaussian mean

from −5 to 5, up to 2 decimal places

Gaussian standard deviation

from 0.1 to 5, up to 2 decimal places

Interval lower bound

from −20 to 35, up to 2 decimal places

Interval upper bound

from −20 to 35, up to 2 decimal places

Move the lower interval bound

A PDF height is density, with inverse-value units. Interval probability is area: approximately 0.682689. A single continuous point has probability zero.

f(x)=\frac1{\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)

P(\ell\le X\le u)=\Phi(z_u)-\Phi(z_\ell)\approx 0.682689

Specify a normalized density

The Gaussian density is normalized analytically over the full real line, not over the displayed window.

f(x)=\frac1{\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)

SpeedStep 1 of 3

Compare Uniform(−2, 2) and Normal(0, 1) on interval [−1, 1]. One probability is exactly 1/2; the other is approximately 0.682689492137.

Predict the uniform density from its support width. Switch to normal mode and explain why equal interval widths need not contain equal mass.

Intuition: shape controls area

A uniform distribution assigns the same density across a finite support. Equal-width subintervals inside that support receive equal probabilities. A Gaussian concentrates density near its mean and decreases smoothly into both tails. Equal-width intervals at different locations generally have different probabilities.

Neither density gives positive probability to a single point. A shaded interval has area; a solitary density height describes concentration per unit of value. This distinction remains true at the highest point of the Gaussian curve.

Definition

For $a<b$ , the continuous uniform density is $f(x)=1/(b-a)$ inside $[a,b]$ and zero outside. Its CDF is zero below $a$ , $(x-a)/(b-a)$ within the support, and one above $b$ . Endpoint density choices do not affect any interval probability.

A Gaussian, or normal, distribution with finite mean $\mu$ and positive standard deviation $\sigma$ has density

f(x)=\frac{1}{\sigma\sqrt{2\pi}} \exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).

Its mathematical support is the whole real line. The standard normal has $\mu=0$ and $\sigma=1$ ; write its CDF as $\Phi$ . The density integrates to one over the whole line. A zero-width uniform support or nonpositive Gaussian $\sigma$ is invalid for these continuous formulas; a point-mass model would be a separate definition.

Mechanics: separate model and viewport

Choose uniform or normal mode, change its parameters, and move the interval bounds. The highlighted area, CDF difference, inspector, formula, and integration trace all describe the same state. In uniform mode, only the part of the interval inside the support contributes area.

In normal mode, the drawn curve is sampled most densely over $\mu\pm4\sigma$ , with extra points at interval and viewport bounds. The viewport also includes the uniform support for comparison. These finite samples make a readable approximation; they do not define the distribution’s support or renormalize its CDF. The numerical integration uses a separate positive-tail cutoff at $8$ standard deviations.

Derivation: normalize and standardize

Uniform normalization requires density times support width to equal one. Thus interval probability is the overlap width divided by $b-a$ :

P(\ell\le X\le u) =\frac{\max(0,\min(u,b)-\max(\ell,a))}{b-a},\qquad \ell\le u.

For the Gaussian, set $Z=(X-\mu)/\sigma$ . The substitution $dx=\sigma\,dz$ cancels the density’s $1/\sigma$ factor, yielding the standard-normal density. Therefore

P(\ell\le X\le u) =\Phi\!\left(\frac{u-\mu}{\sigma}\right) -\Phi\!\left(\frac{\ell-\mu}{\sigma}\right).

The positive scale preserves endpoint order. Because these models are continuous, including or excluding finite endpoints gives the same probability.

Worked examples

Uniform $(-2,2)$ has support width $4$ and density $1/4$ . Its interval $[-1,1]$ has width $2$ , so the area is exactly $1/2$ . The interval $[-3,3]$ overlaps the full support and has probability $1$ , rather than $6\times1/4$ .

Worked example: Compare flat and Gaussian densities

PDF comparison · solid selected, dashed alternative

x · density has inverse-x units · drawn tails are truncated, total Gaussian mass is analytically one

CDF · accumulated area

F(x) = area to the left · approaches zero and one in the tails

Selected interval area: 0.682689
CDF at lower bound: 0.158655
CDF at upper bound: 0.841345
Density at Gaussian mean: 0.398942
Any exact point probability: 0
Standardized lower bound: −1
Standardized upper bound: 1

= exact model quantity · ≈ numerical companion or observed estimate

Probability representation

Uniform start

from −10 to 10, up to 2 decimal places

Uniform end

from −10 to 10, up to 2 decimal places

Gaussian mean

from −5 to 5, up to 2 decimal places

Gaussian standard deviation

from 0.1 to 5, up to 2 decimal places

Interval lower bound

from −20 to 35, up to 2 decimal places

Interval upper bound

from −20 to 35, up to 2 decimal places

Move the lower interval bound

A PDF height is density, with inverse-value units. Interval probability is area: approximately 0.682689. A single continuous point has probability zero.

f(x)=\frac1{\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)

P(\ell\le X\le u)=\Phi(z_u)-\Phi(z_\ell)\approx 0.682689

Specify a normalized density

The Gaussian density is normalized analytically over the full real line, not over the displayed window.

f(x)=\frac1{\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)

SpeedStep 1 of 3

Reproduce the half-area uniform interval, then switch to the standard normal and compare its one-sigma area. Changing the curve shape changes the probability.

For $X\sim\mathcal N(0,1)$ ,

P(-1\le X\le1)=\Phi(1)-\Phi(-1)\approx0.682689492137.

For $X\sim\mathcal N(2,9)$ , the second parameter here denotes variance, so $\sigma=3$ . Interval $[-1,5]$ standardizes to $[-1,1]$ and has the same probability. Always state whether a normal notation or software API uses variance or standard deviation as its scale parameter.

Executable reference code

The exact uniform reference uses Fraction overlap widths. Its independent Gaussian CDF uses standard-library math.erfc, with $\Phi(z)=\tfrac12\operatorname{erfc}(-z/\sqrt2)$ . It also demonstrates the positive-tail Simpson calculation and compares it with that independent CDF over a deterministic sweep and several intervals.

The NumPy program exposes density arrays, an evenly spaced 2049-point grid, and Simpson weights $1,4,2,4,\ldots,2,4,1$ . It uses installed NumPy plus the standard library, without SciPy. Both references include assertions for literal areas, invalid parameters, and the stated CDF and interval error budgets.

code/python/probability_continuous.py

"""Exact uniform areas and Gaussian erfc cross-checks of Simpson integration.

The Gaussian drawing stops at +/-4 sigma. Integration uses the positive tail
to 8 sigma and 2048 Simpson subdivisions. Omitted one-sided mass is <6.4e-16.
CDF absolute-error budget: 1e-9; two-CDF interval budget: 2e-9.
"""
from fractions import Fraction as F
from math import erfc, exp, fsum, isfinite, pi, sqrt


def uniform_interval(a, b, lower, upper):
    if any(not isinstance(x, (int, F)) or isinstance(x, bool) for x in (a, b, lower, upper)):
        raise ValueError("Use integer or Fraction endpoints")
    if a >= b or lower > upper:
        raise ValueError("Require a < b and lower <= upper")
    overlap = max(F(0), min(F(b), F(upper)) - max(F(a), F(lower)))
    return overlap / (F(b) - F(a))


def standardize(x, mu=0, sigma=1):
    if not all(isfinite(v) for v in (x, mu, sigma)) or sigma <= 0:
        raise ValueError("Use finite x, mu and positive finite sigma")
    return (x - mu) / sigma


def normal_cdf(x, mu=0, sigma=1):
    return 0.5 * erfc(-standardize(x, mu, sigma) / sqrt(2))


def simpson_cdf(x, mu=0, sigma=1):
    z = standardize(x, mu, sigma)
    start = abs(z)
    if start >= 8:
        return 1.0 if z > 0 else 0.0
    n = 2048
    h = (8 - start) / n
    def density(t):
        return exp(-t*t/2) / sqrt(2*pi)
    odd = fsum(density(start + j*h) for j in range(1, n, 2))
    even = fsum(density(start + j*h) for j in range(2, n, 2))
    tail = h/3 * (density(start) + density(8) + 4*odd + 2*even)
    return 1 - tail if z >= 0 else tail


def normal_interval(lower, upper, mu=0, sigma=1):
    low, high = normal_cdf(lower, mu, sigma), normal_cdf(upper, mu, sigma)
    if lower > upper:
        raise ValueError("Interval endpoints must be ordered")
    return high - low


def main():
    assert uniform_interval(-2, 2, -1, 1) == F(1, 2)
    assert uniform_interval(-2, 2, -3, 3) == 1
    assert uniform_interval(-2, 2, 2, 3) == 0
    assert uniform_interval(-2, 2, 0, 0) == 0
    assert standardize(-1, 2, 3) == -1 and standardize(5, 2, 3) == 1
    assert abs(normal_interval(-1, 1) - 0.682689492137) < 1e-12
    assert normal_cdf(0) == 0.5
    assert 0.5 * erfc(8 / sqrt(2)) < 6.4e-16
    points = [j/8 for j in range(-80, 81)]
    errors = [abs(simpson_cdf(z) - normal_cdf(z)) for z in points]
    assert max(errors) <= 1e-9
    for lower, upper in ((-1, 1), (-4, 4), (-8, -7), (7, 8), (-2.3, 0.7), (0, 0)):
        estimate = simpson_cdf(upper) - simpson_cdf(lower)
        assert abs(estimate - normal_interval(lower, upper)) <= 2e-9
    assert abs(normal_interval(-1, 5, 2, 3) - 0.682689492137) < 1e-12
    for args in ((0, 0, 0), (0, 0, -1), (float('nan'), 0, 1), (0, float('inf'), 1)):
        try:
            normal_cdf(*args)
        except ValueError:
            pass
        else:
            raise AssertionError("Invalid Gaussian parameters accepted")
    for args in ((2, 2, 0, 1), (-2, 2, 1, -1)):
        try:
            uniform_interval(*args)
        except ValueError:
            pass
        else:
            raise AssertionError("Invalid uniform parameters accepted")
    try:
        normal_interval(1, -1)
    except ValueError:
        pass
    else:
        raise AssertionError("Reversed Gaussian interval accepted")
    print("Uniform[-2,2] density = 1/4, P([-1,1]) =", uniform_interval(-2, 2, -1, 1))
    print("Standard normal P([-1,1]) ≈", normal_interval(-1, 1))
    print("Largest sampled Simpson CDF error against erfc:", max(errors))
    print("Omitted drawn +/-4 sigma tail mass ≈", erfc(4 / sqrt(2)))


if __name__ == "__main__":
    main()

code/python/probability_continuous_numpy.py

"""Array mechanics for densities and positive-tail Simpson, without SciPy."""
from math import erfc, sqrt
import numpy as np


def normal_pdf(x, mu=0, sigma=1):
    x = np.asarray(x, dtype=float)
    if not np.all(np.isfinite(x)) or not np.isfinite(mu) or not np.isfinite(sigma) or sigma <= 0:
        raise ValueError("Use finite points, finite mu and positive finite sigma")
    z = (x - mu) / sigma
    return np.exp(-z*z/2) / (sigma * np.sqrt(2*np.pi))


def simpson_cdf(x, mu=0, sigma=1):
    if not all(np.isfinite(v) for v in (x, mu, sigma)) or sigma <= 0:
        raise ValueError("Use finite x, mu and positive finite sigma")
    z = (x - mu) / sigma
    start = abs(z)
    if start >= 8:
        return 1.0 if z > 0 else 0.0
    grid = np.linspace(start, 8, 2049)
    weights = np.ones(2049)
    weights[1:-1:2] = 4
    weights[2:-1:2] = 2
    tail = ((8 - start) / 2048) / 3 * np.dot(weights, normal_pdf(grid))
    return 1 - tail if z >= 0 else tail


def uniform_interval(a, b, lower, upper):
    if not all(np.isfinite(v) for v in (a, b, lower, upper)) or a >= b or lower > upper:
        raise ValueError("Require finite ordered endpoints and a < b")
    return max(0, min(b, upper) - max(a, lower)) / (b - a)


def main():
    assert uniform_interval(-2, 2, -1, 1) == 0.5
    assert uniform_interval(-2, 2, -3, 3) == 1
    assert uniform_interval(-2, 2, 2, 3) == 0
    np.testing.assert_allclose(normal_pdf([0, 1, -1]), [0.3989422804014327, 0.24197072451914337, 0.24197072451914337], rtol=0, atol=1e-12)
    points = np.linspace(-10, 10, 161)
    estimates = np.array([simpson_cdf(float(z)) for z in points])
    independent = np.array([0.5 * erfc(-float(z) / sqrt(2)) for z in points])
    np.testing.assert_allclose(estimates, independent, rtol=0, atol=1e-9)
    assert np.all(np.diff(estimates) >= 0)
    for lower, upper in ((-1, 1), (-4, 4), (-8, -7), (7, 8), (-2.3, .7), (0, 0)):
        area = simpson_cdf(upper) - simpson_cdf(lower)
        oracle = 0.5 * (erfc(-upper/sqrt(2)) - erfc(-lower/sqrt(2)))
        np.testing.assert_allclose(area, oracle, rtol=0, atol=2e-9)
    np.testing.assert_allclose(simpson_cdf(1) - simpson_cdf(-1), 0.682689492137, rtol=0, atol=2e-9)
    np.testing.assert_allclose(simpson_cdf(5, 2, 3) - simpson_cdf(-1, 2, 3), 0.682689492137, rtol=0, atol=2e-9)
    for args in ((0, 0, 0), (0, 0, -1), (np.nan, 0, 1), (0, np.inf, 1)):
        for function in (normal_pdf, simpson_cdf):
            try:
                function(*args)
            except ValueError:
                pass
            else:
                raise AssertionError("Invalid Gaussian parameters accepted")
    for args in ((2, 2, 0, 1), (-2, 2, 1, -1), (-2, 2, np.nan, 1)):
        try:
            uniform_interval(*args)
        except ValueError:
            pass
        else:
            raise AssertionError("Invalid uniform parameters accepted")
    print("Gaussian interval [-1,1] ≈", simpson_cdf(1) - simpson_cdf(-1))
    print("Largest sampled CDF error:", np.max(np.abs(estimates - independent)))
    print("CDF budget 1e-9; two-CDF interval budget 2e-9; plotted +/-4σ omits tails")


if __name__ == "__main__":
    main()

Explore

In uniform mode, move a fixed-width interval entirely within the support. Predict the unchanged area, then move it across an edge and predict the decrease. Collapse the interval bounds to one point: area becomes zero while the density at that point can remain positive.

In normal mode, compare intervals one standard deviation wide near the mean and far into a tail. Change $\mu$ while shifting the interval by the same amount. Then scale $\sigma$ and the interval’s distances from $\mu$ together. Standardized bounds explain why those coordinated changes preserve probability.

Common mistakes and counterexamples

Uniform means equal density within the support, not over the entire real line. Extending a constant positive height forever would give infinite area. Gaussian does not mean every symmetric interval has the same probability: its location and width relative to $\mu$ and $\sigma$ matter.

For a normal density, substituting variance for $\sigma$ changes both height and width. A curve drawn within four standard deviations also does not capture exactly all mass: about $0.000063342484$ lies beyond the two drawing limits in total.

Numerical considerations

The Gaussian model integrates the positive tail from $|z|$ to $8$ using composite Simpson’s rule with 2048 subdivisions, then uses symmetry to obtain a CDF value. Beyond $8$ standard deviations it truncates the one-sided tail approximately; that omitted mass is less than $6.4\times10^{-16}$ , not mathematically zero.

The absolute CDF error budget is $10^{-9}$ against independent math.erfc values. An interval subtracts two approximate CDFs, so its absolute budget is $2\times10^{-9}$ . This is not a promise of relative accuracy for tiny intervals or extreme tails. The $\pm4\sigma$ drawn curve is not claimed to have exactly normalized area.

Engineering and ML connections

Uniform models describe deliberate random choices over bounded ranges. Gaussian models describe a specified noise or uncertainty shape with location and scale. A bell-shaped drawing alone does not establish that observed data are Gaussian. The interval calculation is conditional on the chosen model, and density values must be interpreted with their measurement units.

Exercises

Use exact fractions for uniform calculations and signed numbers for standardized endpoints. The Gaussian choices use approximate values explicitly.

Summary and checkpoints

Uniform probability is normalized overlap width; Gaussian probability is a standardized CDF difference. Check that you can obtain the uniform half-area exactly, standardize $[-1,5]$ under $\mu=2$ , $\sigma=3$ , and explain both the numerical error budgets and the positive tails beyond a finite drawing.

What changes when probability is spread evenly or concentrated near a center?

Compare flat and Gaussian densities

Intuition: shape controls area

Definition

Mechanics: separate model and viewport

Derivation: normalize and standardize

Worked examples

Worked example: Compare flat and Gaussian densities

Executable reference code

Explore

Common mistakes and counterexamples

Numerical considerations

Engineering and ML connections

Exercises

Measure uniform area

Intersect the interval with the support

Translate a Gaussian interval

Predict a one-sigma probability

Distinguish a viewport from the distribution

Summary and checkpoints

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