Continuous random variable

A random variable X, taking values in R, is said to be continuous if there exists a function fx: R–>[0,+∞) such that for every x ϵ R Fx(x) = ∫ fx(t) dt The function fx is called the density of X

Let f: R–> [0,+∞) such that ∫ f(x) dx = 1 Then there exists a random variable X such that fx = f

▪ Fx is continuous. This implies P(X=x) = 0 ∀x ϵ R ▪ If A = [a, b] P(a<X<b) = ₐ∫ᵇ f(x) dx Moreover P(a<X<b) = ₐ∫ᵇ f(x) dx = = -∞∫ᵇ f(x) dx - -∞∫ᵃ f(x) dx

A sufficient condition is the following: ▪ Fx is continuous. ▪ Fx is differentiable up to a finite number of “exceptionable point” ▪ Except at the exceptional points, F’x is continuous If these properties are satisfied then X is a continuous random variable and fx(x) is equal to • F’x(x) if x isn’t an exceptional point • any value >0 if x an exceptional point

E(X) = ∫ x fx(x) dx g: R –> R E(g(x)) = ∫ g(x) fx(x) dx

Let a<b A random variable X is called Uniform in [a,b] and we write X~U(a,b) or X~Unif(a,b) fx(x) Is equal to • 0 if x !ϵ [a,b] • 1/(b-a) if x ϵ [a,b]

We say that X is an exponential random variable with parameter λ>0, and we write X~Exp(λ) fx(x) is equal to: • λ e^(-λx) for x>0 • 0 for x<0 E(X) = ∫ x fx(x) dx = ∫ x λ e^(-λx) dx E(X) = 1/λ Var(X) = E(X²) - (E(X))² = 1/λ²

A random variable Z is called a standard normal if: fz(z) = (1/ √(2π))*e^(-z²/2) Fz(z) = -∞∫ (1/ √(2π))*e^(-z²/2) dz = Ф(z) Z~N(0,1) E(Z) = 0 Var(Z) = 1

Suppose Z is a standard normal, µ ϵ R, T>0, X = TZ + µ We have a normal random variable if: ▪ X~N( µ ; T² ) ▪ fx(x) = (1/√(2πT²))exp[(1/2T²)(x-µ)²] ▪ E(X) = µ ▪ Var(X) = T² ((X-µ)/T) ~ N(0,1)