University:
York University
Course:
MATH 1131 | Introduction to Statistics I
Academic year:

2022
Views:

459

Pages:

5
Author:

Niklas

Lecture 23 Expected value and variance of continuous random variable Let f(x) be the probability density function of a continuous random variable X. E(x) = ∫xf(x)dx, E(h(x)) = ∫h(x)f(x)dx, Var(x) = E(x²) - (E(x))², Var(ax+b) = a²Var(x) Moreover, the following properties hold (using the properties of the integrals): E(ax+b) = aE(x) + b, Var(x) = E(x²) - (E(x))², Var(ax+b) = a²Var(x) Uniform distribution expected value and variance Let X be a random variable following the Uniform distribution between a and b, thus f(x) = 1/(b-a) if a ≤ x ≤ b 0 otherwise X ~ Unif(a, b) E(x) = ∫xf(x)dx = ∫(x/(b-a))dx from a to b = (b+a)/2 E(x²) = ∫(x²/(b-a))dx = (b²+a²+ab)/(3(b-a)) Var(x) = E(x²) - (E(x))² = (b²-a²)/12 Exponential distribution expected value and variance Let X be a random variable following the Exponential distribution with rate parameter λ: f(x) = λe^(-λx) for x ≥ 0 E(x) = ∫xf(x)dx = ∫(xλe^(-λx))dx = 1/λ Var(x) = E(x²) - (E(x))² = 1/λ² Integration by parts: ∫u dv = uv - ∫v du E(x): ∫xe^(-λx) dx = -(x/λ)e^(-λx) - (1/λ)∫e^(-λx) dx = -(x/λ)e^(-λx) + (1/λ²)e^(-λx) Evaluating from 0 to ∞: (0 - 0) - (0 - 1/λ) = 1/λ E(x²): ∫x²e^(-λx) dx = -(x²/λ)e^(-λx) + (2/λ)∫xe^(-λx) dx Using the result from E(x): -(x²/λ)e^(-λx) + (2/λ) * (1/λ) Evaluating from 0 to ∞: (0 + 0) - (0 + 2/λ²) = 2/λ² Var(x): Var(x) = E(x²) - (E(x))² = (2/λ²) - (1/λ)² = 1/λ² Median of a continuous random variable: The median of a continuous random variable X with pdf f(x) is the value med(x) or x₀ such that F(med(x)) = ∫_{-∞}^{med(x)} f(x) dx = 0.5 How to find x₀ for a uniform distribution: F(x₀) = (x₀ - a) / (b - a) = 0.5 x₀ = (a + b) / 2 = E(x) Uniform Distribution Quantiles Let X be a continuous random variable following uniform distribution between a and b, then: F(x) = (x - a) / (b - a) xq = F^(-1)(q) = a + q(b - a) Exponential Distribution Quantiles Let X be a continuous random variable following exponential distribution with rate parameter λ > 0, then: F(x) = 1 - e^(-λx) xq = F^(-1)(q) = -ln(1 - q) / λ