Named Continuous Distributions

You'll meet five named continuous families — uniform, exponential, gamma, normal, and beta — and learn which real-world question each one answers.

By the end you'll be able to match a waiting-time, magnitude, or proportion scenario to the right continuous family, read its mean and variance off the reference table, and connect the exponential/gamma/Poisson trio the exam leans on.

Predict: if you widen the gap between the lower and upper bound, what happens to the shaded probability — and to the height of the curve in between? Then drag the sliders below to check.

Pick a family, tune its parameters, then set the lower/upper bounds to shade \(P(a \le X \le b)\) — computed by numeric integration under the density curve. Hover the curve for the exact density \(f(x)\) at a point.

Mean = — Variance = — P(a ≤ X ≤ b) = —
Density f(x) — shaded region is the selected probability

You'll use five named continuous families — uniform, exponential (memoryless), gamma, normal, and beta — to model waiting times, magnitudes, and proportions. The exponential/gamma/Poisson relationships below come up constantly on the exam, so lean on them.

Intuitive

Think of Uniform as "equally likely anywhere in a range." Exponential is "how long until the next rare event" — with the odd property that it doesn't matter how long you've already waited, the clock resets. Gamma is "how long until the αth rare event." Normal is the familiar bell curve you'd reach for to model naturally-varying magnitudes. And Beta is a flexible shape confined to [0,1], perfect when you're modeling a proportion.

Formal

Check the reference table below for each family's pdf, mean, and variance. Here's the memoryless property stated precisely: for Exponential, \(P(X>s+t \mid X>s) = P(X>t)\) — unique among continuous distributions. The sum of \(\alpha\) iid Exponential(\(\lambda\)) waits is Gamma(\(\alpha,\lambda\)). If events follow a Poisson process with rate \(\lambda\), the count in a fixed window is Poisson(\(\lambda t\)) while the wait to the \(n\)th event is Gamma(\(n,\lambda\)) — two views of the same process that have to agree.

Applied

Say you're an actuary: the time between insurance claims arriving to an adjuster's desk is often modeled Exponential(\(\lambda\)), so the time until the 3rd claim arrives is Gamma(3,\(\lambda\)). You'd model loss severity with a Gamma or Lognormal-family curve. And if you need the fraction of a policy limit actually paid out — a proportion in [0,1] — Beta is the natural fit.

Here's your cheat sheet — memorize the pdf shape and the mean/variance formulas, and note each family's support (the x-range where f(x) > 0).

Named continuous distributions — pdf, mean, variance
Distribution pdf f(x) Mean Variance
Uniform(a,b) 1/(b−a), a≤x≤b (a+b)/2 (b−a)²/12
Exponential(λ) λe^{−λx}, x≥0 1/λ 1/λ²
Gamma(α,λ) λᵅxᵅ⁻¹e^{−λx}/Γ(α), x≥0 α/λ α/λ²
Normal(μ,σ²) (1/(σ√(2π)))·e^{−(x−μ)²/(2σ²)} μ σ²
Beta(a,b) xᵃ⁻¹(1−x)ᵃ⁻¹/B(a,b), 0≤x≤1 a/(a+b) ab/[(a+b)²(a+b+1)]
Worked example

Calls arrive Poisson(λ=2/hr). Let T = time to the 3rd call. Since T is the wait to the \(n=3\)rd event of a Poisson(2) process, \(T \sim\) Gamma(3,2). Its mean is \(E[T] = \alpha/\lambda = 3/2 = \) 1.5 hours, and its variance is \(\mathrm{Var}(T) = \alpha/\lambda^2 = 3/4 = 0.75\).

Your turn

Same call center, same rate λ=2/hr. Now let \(T_5\) = the time until the 5th call. You already know the pattern: waiting for the \(n\)th event of a Poisson(λ) process is Gamma(\(n,\lambda\)).

So \(T_5 \sim\) Gamma(____, ____), with mean \(E[T_5] = \alpha/\lambda = \) ____.

Reveal the answer

\(T_5 \sim\) Gamma(5, 2), so \(E[T_5] = 5/2 = \) 2.5 hours. Try it in the interactive above: pick Gamma, set shape α = 5 and rate λ = 2, and check the readout Mean matches.

More info — why exponential, gamma, and Poisson share a family

Picture the same stream of rare events (claims, calls, radioactive decays) from two angles. Count events in a fixed window and you get the discrete Poisson(\(\lambda t\)) distribution. Ask instead "how long until the next event?" and you get the continuous Exponential(\(\lambda\)). Ask "how long until the αth event?" and the answer is Gamma(\(\alpha,\lambda\)) — literally a sum of α independent Exponential(\(\lambda\)) waits stacked end to end. It's one underlying process, described with three different distributions depending on what you're asking. The Seeing Theory link in Dive deeper below lets you compare these shapes interactively.

Check your understanding

Question 1 of 4

Calls arrive as a Poisson process at rate λ=2/hour. T = time until the 3rd call. What is the distribution and mean of T?

Question 2 of 4

X has support [0,2]. Which of these could NOT be a valid pdf for X: (A) f(x)=1/2, (B) f(x)=x/2, (C) f(x)=1−x/2, (D) f(x)=x−1?

Question 3 of 4

A call center's calls form a Poisson process at λ=4 calls/hour (the discrete count-in-a-window distribution). What distribution describes the waiting time until the very next call, and what's its mean?

Question 4 of 4

X ~ Uniform(2,10). Using linearity of expectation, what is E[3X − 1]?

Recap

  • Uniform(a,b): flat density over [a,b], mean (a+b)/2, variance (b−a)²/12.
  • Exponential(λ): memoryless waiting time to the next rare event, mean 1/λ, variance 1/λ².
  • Gamma(α,λ): waiting time to the αth rare event — a sum of α iid Exponential(λ) waits — mean α/λ, variance α/λ².
  • Normal(μ,σ²): the bell curve for naturally-varying magnitudes.
  • Beta(a,b): flexible shape on [0,1] for modeling proportions.
  • If events form a Poisson(λ) process, counts in a window are Poisson(λt) while the wait to the nth event is Gamma(n,λ) — the same process seen two ways.

Dive deeper

Sources

  • Named Continuous Distributions