Continuous expectation
Objective. Calculate moments from density functions.
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- 2. Worked example
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Learn: the key idea
For a continuous random variable with density function f(x), the expectation (mean) is E(X) = ∫xf(x)dx over the domain, and more generally E(g(X)) = ∫g(x)f(x)dx for any function g. The variance is Var(X) = E(X²) − [E(X)]², requiring the separate calculation of E(X²) = ∫x²f(x)dx. These moments summarise the centre and spread of the distribution, directly analogous to the discrete formulae but using integration instead of summation. For example, for a uniform density on [0,2], E(X) = ∫₀^2 x(1/2)dx = 1, the midpoint, as expected by symmetry.
Worked example
A continuous random variable X has density f(x) = 3x² for 0 ≤ x ≤ 1. Find E(X) and Var(X).
- 1E(X) = ∫₀^1 x(3x²)dx = ∫₀^1 3x³ dx = [3x⁴/4]₀^1 = 3/4
- 2E(X²) = ∫₀^1 x²(3x²)dx = ∫₀^1 3x⁴ dx = [3x⁵/5]₀^1 = 3/5
- 3Var(X) = E(X²) − [E(X)]² = 3/5 − (3/4)²
- 4Simplify: 3/5 − 9/16 = 48/80 − 45/80 = 3/80
Answer: E(X) = 3/4, Var(X) = 3/80
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Start practisingCommon mistake. Students compute Var(X) as E((X−μ)²) by expanding incorrectly, or forget to square E(X) separately, writing Var(X) = E(X²) − E(X). Always compute E(X²) and [E(X)]² separately before subtracting.
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