Expectation and Variance of Continuous Distributions

This lesson focuses on computing $E(X)$E(X), $E(X^2)$E(X2), $\text{Var}(X)$Var(X), and related quantities for continuous distributions. You will practise integration techniques and learn how to handle more complex PDFs involving polynomials, trigonometric functions, and parameters.

---

Core Formulas (Recap)

For a continuous random variable $X$X with PDF $f(x)$f(x) defined on $[a, b]$[a,b]:

Quantity	Formula
$E(X)$E(X)	$\int_a^b x f(x) \, dx$∫ab​xf(x)dx
$E(X^2)$E(X2)	$\int_a^b x^2 f(x) \, dx$∫ab​x2f(x)dx
$E(g(X))$E(g(X))	$\int_a^b g(x) f(x) \, dx$∫ab​g(x)f(x)dx
$\text{Var}(X)$Var(X)	$E(X^2) - (E(X))^2$E(X2)−(E(X))2

---

Worked Example 1: Polynomial PDF

Let $f(x) = 12x^2(1 - x)$f(x)=12x2(1−x) for $0 \leq x \leq 1$0≤x≤1.

Verify: $\int_0^1 12x^2(1-x) \, dx = 12\int_0^1 (x^2 - x^3) \, dx = 12\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 12\left(\frac{1}{3} - \frac{1}{4}\right) = 12 \times \frac{1}{12} = 1$∫01​12x2(1−x)dx=12∫01​(x2−x3)dx=12[3x3​−4x4​]01​=12(31​−41​)=12×121​=1. Valid.

Mean:

$E(X) = \int_0^1 12x^3(1-x) \, dx = 12\int_0^1 (x^3 - x^4) \, dx = 12\left[\frac{x^4}{4} - \frac{x^5}{5}\right]_0^1 = 12\left(\frac{1}{4} - \frac{1}{5}\right) = 12 \times \frac{1}{20} = \frac{3}{5} = 0.6$E(X)=∫01​12x3(1−x)dx=12∫01​(x3−x4)dx=12[4x4​−5x5​]01​=12(41​−51​)=12×201​=53​=0.6

Second moment: