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07-01: Continuous, Uniform and Exponential Distributions

Discrete variables have gaps; continuous variables do not. That one change forces a different way of computing probability — areas instead of sums — and it is the setup for the normal distribution in 07-02.

From Sums to Areas

For a discrete variable you add probabilities. For a continuous variable you measure the area under a curve.

Discrete       P(X = x) is a real number, and Σ P(x) = 1

Continuous     P(X = x) = 0  for every single value x
               P(a ≤ X ≤ b) = area under the density curve between a and b
               Total area under the curve = 1

Why P(X = x) = 0? There are infinitely many possible values, so no single one carries positive probability. Ask instead for the probability of an interval.

A convenient consequence: with continuous variables the endpoints do not matter.

P(X < a)  =  P(X ≤ a)          P(a < X < b)  =  P(a ≤ X ≤ b)

That is not true for discrete variables — a distinction worth being deliberate about.

Density function vs. distribution function

Function Symbol Meaning
Probability density function (PDF) f(x) Height of the curve. Not a probability — it can exceed 1
Cumulative distribution function (CDF) F(x) = P(X ≤ x) Area to the left of x. Always between 0 and 1

Everything reduces to CDF arithmetic:

P(X ≤ b)      =  F(b)
P(X > a)      =  1 − F(a)
P(a < X < b)  =  F(b) − F(a)

Every software function in this chapter is one of those three lines.

The Continuous Uniform Distribution

All values in [a, b] are equally likely — the flat rectangle.

             1
f(x)  =  ───────         for  a ≤ x ≤ b,   0 otherwise
           b − a

             x − a
F(x)  =  ───────────     for  a ≤ x ≤ b
             b − a

                         d − c
P(c ≤ X ≤ d)  =  ─────────────────       for a ≤ c ≤ d ≤ b
                         b − a

                a + b                      (b − a)²
Mean   μ  =  ───────────      Variance σ² = ──────────
                  2                            12

Worked example

A bus arrives at a uniformly random time between 0 and 20 minutes after you arrive.

a = 0,  b = 20,  f(x) = 1/20 = 0.05

P(wait ≤ 5 min)        = (5 − 0)/20  = 0.25
P(wait between 8 and 12) = (12 − 8)/20 = 0.20
P(wait > 15)           = (20 − 15)/20 = 0.25

μ = (0 + 20)/2 = 10 minutes
σ = √(20²/12) = √33.33 = 5.77 minutes

The Exponential Distribution

Models the waiting time between events that occur at a constant rate — the continuous partner of the Poisson distribution of 06-02. If events arrive Poisson with rate λ per unit time, the gaps between them are Exponential with the same λ.

f(x)  =  λ · e^(−λx)              x ≥ 0

F(x)  =  1 − e^(−λx)              P(X ≤ x)

P(X > x)  =  e^(−λx)              the survival function

                1                            1                    1
Mean   μ  =  ─────      Variance  σ² =  ─────       SD  σ  =  ─────
                λ                          λ²                    λ

Mean and standard deviation are equal — the exponential is always strongly right-skewed.

Memorylessness: P(X > s + t | X > s) = P(X > t). A component that has already survived 5 years is, under this model, exactly as likely to survive another year as a brand-new one.

Worked example

Customers arrive at a rate of 4 per hour, so λ = 4 per hour (mean gap 1/4 hour = 15 minutes).

P(next arrival within 10 min)  = P(X ≤ 1/6 hr) = 1 − e^(−4/6)   = 1 − 0.5134 = 0.4866
P(next arrival takes > 30 min) = P(X > 0.5 hr) = e^(−4 × 0.5)   = e^(−2)     = 0.1353
P(gap between 10 and 30 min)   = 0.8647 − 0.4866 = 0.3781

μ = 1/4 hour = 15 minutes,   σ = 15 minutes

Cross-check with the Poisson. P(no arrivals in 30 minutes) from the Poisson with λ = 4(0.5) = 2 is e^(−2) = 0.1353 — the same number. Two views of one process.

Excel

' ── UNIFORM (no built-in — the formulas are one-liners) ─────────────
' a in B1, b in B2
=1/(B2-B1)                      ' density f(x)                 -> 0.05
=(5-B1)/(B2-B1)                 ' P(X ≤ 5)                     -> 0.25
=(12-8)/(B2-B1)                 ' P(8 ≤ X ≤ 12)                -> 0.20
=1-(15-B1)/(B2-B1)              ' P(X > 15)                    -> 0.25
=(B1+B2)/2                      ' mean                         -> 10
=SQRT((B2-B1)^2/12)             ' standard deviation           -> 5.7735
=B1+RAND()*(B2-B1)              ' one random uniform draw

' ── EXPONENTIAL ─────────────────────────────────────────────────────
=EXPON.DIST(1/6, 4, TRUE)       ' P(X ≤ 1/6 hr)  CDF           -> 0.48658
=EXPON.DIST(1/6, 4, FALSE)      ' density f(1/6)               -> 2.05366
=1-EXPON.DIST(0.5, 4, TRUE)     ' P(X > 0.5 hr)                -> 0.13534
=EXP(-4*0.5)                    ' same, by formula             -> 0.13534
=EXPON.DIST(0.5,4,TRUE)-EXPON.DIST(1/6,4,TRUE)   ' P(1/6 < X < 0.5) -> 0.37808
=1/4                            ' mean and SD (both = 1/λ)     -> 0.25 hr

' Inverse: the time by which 90% of gaps have elapsed
=-LN(1-0.9)/4                   ' -> 0.5756 hours ≈ 34.5 minutes

' Random exponential draw
=-LN(RAND())/4

' ── Cross-check against Poisson ─────────────────────────────────────
=POISSON.DIST(0, 4*0.5, FALSE)  ' P(no arrivals in 30 min)     -> 0.13534

Note

Excel's EXPON.DIST(x, lambda, cumulative) takes the rate λ, not the mean. If a problem gives you "a mean of 15 minutes", convert: λ = 1/15 per minute, or λ = 4 per hour. Keep the time units consistent throughout.

R

# ── UNIFORM:  dunif / punif / qunif / runif ────────────────────────
dunif(7, min = 0, max = 20)          # density        -> 0.05
punif(5, 0, 20)                      # P(X ≤ 5)       -> 0.25
punif(12, 0, 20) - punif(8, 0, 20)   # P(8 ≤ X ≤ 12)  -> 0.20
punif(15, 0, 20, lower.tail = FALSE) # P(X > 15)      -> 0.25
qunif(0.90, 0, 20)                   # 90th percentile-> 18
runif(5, 0, 20)                      # 5 random draws

c(mean = (0 + 20)/2, sd = sqrt((20 - 0)^2 / 12))     # 10, 5.7735

curve(dunif(x, 0, 20), from = -2, to = 22, n = 1000,
      col = "#5B2A86", lwd = 2, ylab = "f(x)", main = "Uniform(0, 20)")

# ── EXPONENTIAL:  dexp / pexp / qexp / rexp ────────────────────────
lambda <- 4                                   # arrivals per hour

pexp(1/6, rate = lambda)                      # P(X ≤ 10 min)  -> 0.48658
pexp(0.5, lambda, lower.tail = FALSE)         # P(X > 30 min)  -> 0.13534
pexp(0.5, lambda) - pexp(1/6, lambda)         # P(10 < X < 30) -> 0.37808
qexp(0.90, lambda)                            # 90th pct -> 0.5756 hr
rexp(5, lambda)

c(mean = 1/lambda, sd = 1/lambda)             # 0.25, 0.25 hours

curve(dexp(x, lambda), from = 0, to = 1.5, n = 500,
      col = "#0FA3A3", lwd = 2, ylab = "f(x)",
      main = "Exponential(rate = 4)")

# Cross-check against Poisson
dpois(0, lambda = 4 * 0.5)                    # 0.13534  — identical

# Memorylessness, verified
pexp(6, 1, lower.tail = FALSE) / pexp(5, 1, lower.tail = FALSE)   # P(X>6 | X>5)
pexp(1, 1, lower.tail = FALSE)                                    # P(X>1) — same

Python

import numpy as np
from scipy import stats

# ── UNIFORM  (loc = a, scale = b − a) ──────────────────────────────
U = stats.uniform(loc=0, scale=20)

U.pdf(7)                       # density        -> 0.05
U.cdf(5)                       # P(X ≤ 5)       -> 0.25
U.cdf(12) - U.cdf(8)           # P(8 ≤ X ≤ 12)  -> 0.20
U.sf(15)                       # P(X > 15)      -> 0.25
U.ppf(0.90)                    # 90th percentile-> 18.0
U.rvs(5, random_state=1)       # random draws
U.mean(), U.std()              # 10.0, 5.7735

# ── EXPONENTIAL  (scipy uses SCALE = 1/λ, not the rate!) ───────────
lam = 4                        # arrivals per hour
E = stats.expon(scale=1 / lam)

E.cdf(1/6)                     # P(X ≤ 10 min)  -> 0.48658
E.sf(0.5)                      # P(X > 30 min)  -> 0.13534
E.cdf(0.5) - E.cdf(1/6)        # P(10 < X < 30) -> 0.37808
E.ppf(0.90)                    # 90th pct       -> 0.5756 hr
E.mean(), E.std()              # 0.25, 0.25

# Cross-check against Poisson
stats.poisson.pmf(0, mu=4 * 0.5)      # 0.13534 — identical

# Memorylessness
E1 = stats.expon(scale=1)
E1.sf(6) / E1.sf(5), E1.sf(1)         # equal

# ── Plot both ──────────────────────────────────────────────────────
import matplotlib.pyplot as plt
fig, (a1, a2) = plt.subplots(1, 2, figsize=(10, 3))
xs = np.linspace(-2, 22, 500)
a1.plot(xs, U.pdf(xs), color="#5B2A86"); a1.set_title("Uniform(0, 20)")
xs = np.linspace(0, 1.5, 500)
a2.plot(xs, E.pdf(xs), color="#0FA3A3"); a2.set_title("Exponential(rate=4)")
plt.tight_layout(); plt.show()

Warning

Parameterization traps. Excel's EXPON.DIST and R's dexp/pexp take the rate λ. SciPy's stats.expon takes the scale 1/λ. Passing 4 where SciPy expects 0.25 gives an answer that looks plausible and is completely wrong. Similarly, SciPy's uniform takes loc and scale = b − a, not a and b.

Quick Reference

Task Excel R Python
Uniform density 1/(b-a) dunif(x,a,b) uniform(a, b-a).pdf(x)
Uniform P(X ≤ x) (x-a)/(b-a) punif(x,a,b) uniform(a, b-a).cdf(x)
Uniform percentile a+p*(b-a) qunif(p,a,b) uniform(a, b-a).ppf(p)
Uniform random a+RAND()*(b-a) runif(n,a,b) uniform(a, b-a).rvs(n)
Exponential density EXPON.DIST(x,λ,FALSE) dexp(x,λ) expon(scale=1/λ).pdf(x)
Exponential P(X ≤ x) EXPON.DIST(x,λ,TRUE) pexp(x,λ) expon(scale=1/λ).cdf(x)
Exponential P(X > x) EXP(-λ*x) pexp(x,λ,lower.tail=FALSE) expon(scale=1/λ).sf(x)
Exponential percentile -LN(1-p)/λ qexp(p,λ) expon(scale=1/λ).ppf(p)
Distribution Mean Variance Shape
Uniform (a,b) (a+b)/2 (b−a)²/12 Flat rectangle
Exponential (λ) 1/λ 1/λ² Right-skewed, decaying

Common Mistakes

  • Asking for P(X = 3) on a continuous variable — it is 0. Ask for an interval.
  • Passing the mean where the software wants the rate (or vice versa).
  • Mixing time units: λ per hour with x in minutes.
  • Treating the density f(x) as a probability. For Uniform(0, 0.5) the density is 2 — perfectly legal, and not a probability.
  • Forgetting that a uniform probability is just (length of the sub-interval) / (total length) — no calculus needed.

Exercises: 07-01: Exercises — Continuous, Uniform and Exponential Distributions

⬅️ Previous: 06-02: Binomial and Poisson Distributions ➡️ Next: 07-02: The Normal Distribution and Z-Scores