Free lesson · Integral calculus

Improper integrals, densities and continuous expectations

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Start with the idea

A density assigns probability per unit of the outcome variable.
Integrating it over a region gives probability; weighting it by an outcome before integrating gives an expectation.
This is the continuous version of a probability-weighted sum.

Symbols, units & horizon

X: nonnegative continuous random variable
x: a possible value of X
p(x): density in inverse x-units
λ: positive rate in inverse x-units
E[X]: mean in x-units
∞: infinite upper bound interpreted as a limit
b: finite temporary upper bound
e: exponential base
P(X>b): probability of exceeding b

When and why to use this

Use density integrals to compute continuous probabilities and expected quantities. Check support, normalisation and moment existence before applying a mean-based model.

A continuous density is nonnegative and integrates to one.
Its height is not itself a probability and may exceed one when its units permit it.
A single point has probability zero under a continuous density, but an interval can have positive probability.
An improper integral has an infinite bound or a singularity and is defined by a limit.
Convergence must be checked.
A normalised density can still have an infinite mean or variance, so valid probabilities do not guarantee finite moments.

p (x) = λ e^{- λ x} (x \geq 0), \int_{0}^{\infty} p (x) d x = 1, 𝔼 [X] = \int_{0}^{\infty} x p (x) d x = \frac{1}{λ}

Calculus: derivation and arithmetic

Improper integrals, densities and continuous expectations

Integrate to a finite bound first: $\int_{0}^{b} λ e^{- λ x} d x = [- e^{- λ x}]_{0}^{b} = 1 - e^{- λ b}$ . For $λ > 0$ , this tends to 1 as $b \to \infty$ .
For the first moment choose $u = x$ , $d v = λ e^{- λ x} d x$ . Then $d u = d x$ and $v = - e^{- λ x}$ .
Integration by parts gives $\int_{0}^{b} x λ e^{- λ x} d x = - b e^{- λ b} + \frac{1 - e^{- λ b}}{λ}$ . Both exponential boundary terms vanish in the limit, leaving $E [X] = \frac{1}{λ}$ .
The tail is the complementary probability: $P (X > b) = 1 - (1 - e^{- λ b}) = e^{- λ b}$ . At $λ = 2$ per minute and $b = 1$ minute, the mean is 0.5 minute and the tail is approximately 0.135335.

Work it by hand

If λ=2 per minute, mean waiting time is .5 minute. The probability of waiting longer than 1 minute is exp(−2)≈.135335. This is a distributional assumption, not a universal waiting-time law.

An analogy to remember

Density is like the thickness of material along a strip. Height at one location is not total mass; area under the thickness curve over an interval is.

How this becomes a building block

Expectations, tail losses and risk-neutral payoff prices are weighted integrals. A waiting-time model can be a component of a toy arrival or execution model. The Poisson/exponential assumptions would need empirical checking before describing actual orders.

Python implementation

Self-contained teaching example. Python 3.10+; dependencies and input conventions are shown in the code and notation. Run in your own Python environment.

from math import exp, expm1

def exponential_model(rate,threshold):
    if rate<=0 or threshold<0: raise ValueError("Positive rate, nonnegative threshold required")
    return {"density":rate*exp(-rate*threshold),
            "cdf":-expm1(-rate*threshold),"tail":exp(-rate*threshold),"mean":1/rate}

print(exponential_model(2,1))

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