Improper integrals and continuous probability

What this concept is

An integral to infinity is defined by a limit of finite integrals. A density is nonnegative and integrates to 1 on its support. Its CDF F(t) accumulates density up to t; where density is continuous F′(t)=f(t).

Why it matters here

Continuous probability needs area rather than point height. Tail truncation introduces a separate error even if numerical integration over the retained interval were exact.

Concrete example

For λ>0, f(t)=λ exp(−λt) on t≥0 and zero otherwise integrates to 1 because ∫₀ᵀ f=1−exp(−λT) tends to 1. F(t)=0 for t<0 and 1−exp(−λt) for t≥0. The mean ∫₀∞ tλ exp(−λt)dt=1/λ by parts, using T exp(−λT) tending to zero. For λ=2 per second, P(T>3 seconds)=exp(−6); mean is 0.5 seconds.

Common confusion / misconception

f(t) can exceed 1; its units here are inverse seconds. Point probabilities are zero for a density model. A finite numerical answer over [0,100] does not prove convergence of an integral to infinity.

How to use it

Write support first, prove nonnegativity and normalization, then integrate over the requested event. Split at singularities: cancellation across an undefined point is not ordinary improper convergence. For ∫₁∞1/x dx, ln T diverges; for ∫₁∞1/x² dx, 1−1/T tends to 1. To omit exponential tail mass at most ε with 0<ε<1, choose T≥−ln(ε)/λ.

Check yourself

Why can 1/x on [1,∞) not be normalized by a positive finite constant? Can a valid density have an infinite mean?

Mini drill or application

Normalize c x on [0,2] and find its mean and probability of [1,2]. Answers: c=1/2, mean 4/3, probability 3/4. Then return to the probability module for dependence, inference and model choice; calculus alone does not justify those assumptions.

Read this only if stuck

OpenStax selective support. Use the reading guide for the section sequence and practice contract. Record your reasoning before consulting solutions.