Continuous Distributions

This lesson covers the continuous distributions in the GATE DA syllabus. The first three model data directly; the t and chi-squared distributions appear when we build confidence intervals and tests in the later lessons.

Uniform(a, b)

Every value in [a, b] is equally likely.

• pdf f(x) = 1/(b − a) on [a, b], cdf F(x) = (x − a)/(b − a)
• E[X] = (a + b)/2, Var(X) = (b − a)²/12
• P(c ≤ X ≤ d) = (d − c)/(b − a) for a ≤ c ≤ d ≤ b: probability is proportional to length.

Exponential(λ)

The waiting time until the next event in a Poisson process with rate λ: the time to the next customer, the lifetime of a component that does not age.

• pdf f(x) = λe^(−λx) for x ≥ 0, cdf F(x) = 1 − e^(−λx)
• P(X > t) = e^(−λt)
• E[X] = 1/λ, Var(X) = 1/λ², median = (ln 2)/λ
• Memoryless: P(X > s + t | X > s) = P(X > t). A component that has survived s hours is as good as new. It is the only memoryless continuous distribution.
• If X₁ and X₂ are independent exponentials with rates λ₁ and λ₂, then min(X₁, X₂) is exponential with rate λ₁ + λ₂.

Example. A bulb's lifetime is exponential with mean 2 years (λ = 0.5). P(it lasts more than 3 years) = e^(−1.5) ≈ 0.2231, and its median lifetime is ln 2 / 0.5 ≈ 1.386 years, less than the mean, because the distribution is skewed to the right.

The normal distribution N(μ, σ²)

The bell curve. It describes measurement errors, heights, test scores, and (by the central limit theorem) averages of almost anything.

• pdf f(x) = (1/(σ√(2π))) e^(−(x − μ)²/(2σ²)), symmetric about μ
• E[X] = μ, Var(X) = σ²; mean = median = mode
• If X ~ N(μ, σ²), then aX + b ~ N(aμ + b, a²σ²).
• Sums of independent normals are normal: X + Y ~ N(μ_X + μ_Y, σ²_X + σ²_Y).

Standardising

Z = (X − μ)/σ has the standard normal distribution N(0, 1). Its cdf is written Φ(z), and tables give Φ(z) for z ≥ 0. By symmetry, Φ(−z) = 1 − Φ(z).

The 68-95-99.7 rule: about 68% of values lie within 1σ of the mean, 95% within 2σ (more precisely 1.96σ), and 99.7% within 3σ.

Example. IQ scores are N(100, 15²). P(IQ > 130) = P(Z > 2) = 1 − 0.9772 = 0.0228, and P(85 < IQ < 115) = P(−1 < Z < 1) ≈ 0.6827.

The chi-squared distribution χ²(k)

If Z₁, …, Zₖ are independent standard normals, then Z₁² + … + Zₖ² ~ χ²(k), chi-squared with k degrees of freedom.

• It takes only non-negative values and is skewed to the right.
• E = k, Var = 2k.
• It is used for the variance of a normal sample, (n − 1)S²/σ² ~ χ²(n − 1), and for goodness-of-fit and independence tests.

Student's t distribution t(ν)

If Z ~ N(0, 1) and V ~ χ²(ν) are independent, then T = Z / √(V/ν) ~ t(ν).

• It is symmetric about 0 and bell-shaped, but with heavier tails than the normal: extreme values are more likely.
• Mean 0 (for ν > 1) and variance ν/(ν − 2) (for ν > 2), which is greater than 1.
• As ν → ∞, t(ν) approaches N(0, 1).
• It appears when we standardise a sample mean using the sample standard deviation instead of the true σ: (X̄ − μ)/(S/√n) ~ t(n − 1) for a normal population.

Critical values used in the next lessons (t_α,ν has upper-tail area α):

from math import exp, log, sqrt
from statistics import NormalDist
import random

iq = NormalDist(100, 15)
print(f"P(IQ > 130) = {1 - iq.cdf(130):.4f}")
print(f"P(85 < IQ < 115) = {iq.cdf(115) - iq.cdf(85):.4f}")
z = NormalDist()
print(f"z for upper tail 0.05: {z.inv_cdf(0.95):.3f}; for 0.025: {z.inv_cdf(0.975):.3f}")
print(f"exponential, mean 2: P(X > 3) = {exp(-3 / 2):.4f}, median = {log(2) * 2:.3f}")

# Chi-squared by simulation: the sum of 3 squared standard normals has mean 3 and variance 6
random.seed(7)
samples = [sum(random.gauss(0, 1) ** 2 for _ in range(3)) for _ in range(100000)]
m = sum(samples) / len(samples)
v = sum((s - m) ** 2 for s in samples) / len(samples)
print(f"chi-squared(3) by simulation: mean = {m:.2f}, variance = {v:.2f}")

P(IQ > 130) = 0.0228
P(85 < IQ < 115) = 0.6827
z for upper tail 0.05: 1.645; for 0.025: 1.960
exponential, mean 2: P(X > 3) = 0.2231, median = 1.386
chi-squared(3) by simulation: mean = 2.99, variance = 5.92

Practice questions