Expected Value of a Continuous Random Variable

Many students who have mastered discrete expected value hit a wall when the sum turns into an integral. The concept is the same, but the tool changes. For a continuous random variable, the expected value integral E[X] = ∫ x f_X(x) dx replaces the sum Σ x p(x). The expected value of the pdf is the long-run average, not a prediction for a single trial. You cannot land on exactly 3.14 from a uniform(0,10) distribution, but the mean of a continuous distribution settles at 5 after enough draws. The integral definition, two full examples, the law of the unconscious statistician (LOTUS) for E(g(X)), and the case where the integral diverges and the expected value does not exist are covered.

The definition of expected value for a discrete random variable is E[X] = Σ x_i P(X = x_i). For a continuous random variable, probability lives in intervals, not points. The probability density function f_X(x) gives the density of probability at x, and the total area under f_X(x) equals 1. The continuous analogue is the integral ∫ x f_X(x) dx over the support of X, as defined in Blitzstein & Hwang ch. 5 and OpenStax Introductory Statistics ch. 5. Both sources stress that f_X(x) is nonnegative and integrates to 1. The integral computes the center of mass of the density. If X has a density that is symmetric about μ, then E[X] = μ, provided the integral converges.

When the Support is Finite

If the random variable is bounded between a and b, the integral runs from a to b. If it is unbounded, you integrate from -∞ to ∞, or from 0 to ∞ for a nonnegative variable like the exponential. The expected value integral must be absolutely convergent; if it is not, the expected value does not exist. This is not a technical detail you can ignore. The Cauchy distribution, covered in the final section, shows what happens when the tails are too heavy.

The uniform distribution has density f_X(x) = 1/(b - a) for a ≤ x ≤ b, zero elsewhere. Compute the expected value for this distribution.

E[X] = ∫_a^b x * [1/(b - a)] dx = [1/(b - a)] * (1/2) x^2 |_a^b = [1/(b - a)] * (b^2 - a^2)/2 = (b + a)/2.

The mean of a uniform distribution is the midpoint of the interval. For uniform(0, 10), E[X] = 5. The variance, from Blitzstein & Hwang ch. 5, is (b - a)^2 / 12. For uniform(0,10), Var(X) = 100/12 = 25/3 ≈ 8.33, and the standard deviation is about 2.89. OpenStax Introductory Statistics ch. 5 gives the same result through the definition Var(X) = ∫ (x - μ)^2 f(x) dx = ∫ (x - μ)^2 / (b - a) dx.

What to Do with the Result

The expected value (a+b)/2 is the long-run average. On any single draw you get a number between a and b, never the mean unless it happens to fall there by chance. The mode is every point in the interval; the median is also (a+b)/2. Over many draws, the sample average converges to (a+b)/2 by the Law of Large Numbers.

The exponential distribution with rate λ has density f_X(x) = λ e^{-λ x} for x ≥ 0, zero for x < 0. The mean of the exponential is 1/λ.

E[X] = ∫_0^∞ x λ e^{-λ x} dx. Integrate by parts: let u = x, dv = λ e^{-λ x} dx. Then du = dx, v = -e^{-λ x}. The integral becomes [-x e^{-λ x}]_0^∞ + ∫_0^∞ e^{-λ x} dx = 0 + [ -1/λ e^{-λ x} ]_0^∞ = 1/λ.

For λ = 0.5, the expected value is 2. The variance, from Blitzstein & Hwang ch. 5, is 1/λ^2. Exponential(0.5) has variance 4 and standard deviation 2. The memoryless property means P(X > s+t | X > s) = P(X > t), which is unique to the exponential among continuous distributions. The median is ln(2)/λ, about 1.39 for λ = 0.5, less than the mean because the distribution is skewed right.

Checking Your Work with LOTUS

Suppose you want E[X^2] for the exponential. Use LOTUS for continuous variables: E[g(X)] = ∫ g(x) f_X(x) dx. For g(x) = x^2, E[X^2] = ∫_0^∞ x^2 λ e^{-λ x} dx. Integrate by parts twice or recognize it as the second moment of the exponential: E[X^2] = 2/λ^2. Then Var(X) = E[X^2] - (E[X])^2 = 2/λ^2 - 1/λ^2 = 1/λ^2, matching the variance from the definition.

LOTUS lets you compute the expected value of any function of X without first finding the distribution of g(X). For a continuous random variable, E[g(X)] = ∫ g(x) f_X(x) dx, as stated in Blitzstein & Hwang ch. 5. The same rule applies for variance: Var(X) = E[(X - μ)^2] = ∫ (x - μ)^2 f_X(x) dx, or equivalently E[X^2] - (E[X])^2.

Practical Steps for Variance

• Compute μ = E[X] = ∫ x f_X(x) dx.
• Compute E[X^2] = ∫ x^2 f_X(x) dx.
• Var(X) = E[X^2] - μ^2.

Linearity of expectation holds for continuous variables as it does for discrete: E[aX + b] = aE[X] + b. This is the same property used for binomial expected value and other discrete distributions, but now applied to integrals. If you know E[X] for a distribution, you can find E[2X - 3] without reintegrating: it is 2E[X] - 3.

Example with a Piecewise Density

Let f_X(x) = 2x on [0,1], zero elsewhere. Check that ∫_0^1 2x dx = 1. E[X] = ∫_0^1 x * 2x dx = 2∫_0^1 x^2 dx = 2/3. E[X^2] = ∫_0^1 x^2 * 2x dx = 2∫_0^1 x^3 dx = 1/2. Then Var(X) = 1/2 - (2/3)^2 = 1/2 - 4/9 = 1/18. The standard deviation is about 0.236.

The Cauchy distribution has density f_X(x) = 1/(π(1 + x^2)) for all real x. The integral ∫_{-∞}^∞ |x| f_X(x) dx diverges because the tails are too heavy. More precisely, ∫_{-∞}^∞ x f_X(x) dx does not converge absolutely, so the expected value does not exist. Blitzstein & Hwang ch. 5 explicitly states that the Cauchy has no mean. The median is 0, and the distribution is symmetric about 0, but symmetry alone does not guarantee a finite mean.

What Goes Wrong

Compute ∫_0^R x / (π(1 + x^2)) dx. As R → ∞, the integral behaves like (1/2π) ln(1 + R^2), which diverges. The positive and negative tails are balanced in principle, but the divergence is too slow. The sample mean of Cauchy draws does not converge to a fixed number; it jumps around with every new observation. If you use a Cauchy model to represent a real phenomenon (e.g., some physical measurements, certain financial returns), you cannot report an expected value as a summary. The mode, median, and interquartile range are still meaningful.

How to Detect a Nonexistent Expected Value

Check whether ∫ |x| f_X(x) dx converges over the support. If the density decays slower than 1/|x|^2 as |x| → ∞, the tail integral diverges. The Cauchy decays like 1/x^2, which is exactly the boundary, it barely fails. The Pareto distribution with shape parameter α ≤ 1 also has no finite mean. If you are working with real data and the histogram shows extreme outliers without a central tendency, test whether a distribution with a nonexistent mean might fit better.

Distribution	Support	Density f(x)	E[X]
Uniform(a,b)	a ≤ x ≤ b	1/(b − a)	(a + b)/2
Exponential(λ)	x ≥ 0	λ e^{-λ x}	1/λ
Normal(μ, σ²)	−∞ < x < ∞	1/(σ √(2π)) e^{-(x−μ)²/(2σ²)}	μ
Gamma(α, λ)	x ≥ 0	λ^α x^{α−1} e^{-λ x} / Γ(α)	α/λ
Beta(α, β)	0 ≤ x ≤ 1	x^{α−1} (1−x)^{β−1} / B(α,β)	α/(α+β)
Cauchy(x₀, γ)	−∞ < x < ∞	γ / (π(γ² + (x−x₀)²))	does not exist

The calculus-based probability student who needs to compute an expected value integral, apply LOTUS, and calculate variance for continuous distributions will find the worked integrals above directly useful. This includes anyone studying from Blitzstein & Hwang or OpenStax Introductory Statistics. The material also suits project managers who use Expected Monetary Value (EMV) in decision trees, since EMV for continuous risk impacts involves integrating over probability distributions. Casual users who want to check a single lottery EV number should use a discrete calculator; continuous expected value applies when the outcome can take any value in an interval, not a countable set of prizes. If you need a guarantee about a single trial, or if you are looking for a specific outcome probability rather than the long-run average, this content is not for you. The single thing that most often goes wrong is treating the integral as optional: if the tails are heavy, the integral diverges, and reporting a finite mean is a mistake that misleads every decision you base on it.