Back to marketplace

Exponential distribution

No description yet. This curated deck is ready for daily study.

Total cards: 7•Content version: v4•Public deck•Updated: 8/1/2026

Card preview (7)

#1
Front (Question)

Exponential distribution ▪ Definition

Back (Answer)

A continuous random variable X is said to have Exponential distribution with parameter λ>0 X ~ Exp(λ) Density: fₓ(x)= ▪ λ*e^(-λx) if x⩾0 ▪ 0 if x<0 Fₓ(x)= ▪ 1-e^(-λx) if x⩾0 ▪ 0 if x<0 Then E(X) = 1/λ Var(X) = 1/λ²

#2
Front (Question)

Exponential distribution ▪ When use it

Back (Answer)

▪ time before earthquake ▪ time before customer enter in a shop ▪ time next call ▪ time piece of machinery work

#3
Front (Question)

Exponential distribution ▪ Properties

Back (Answer)

a) Loss of memory b) Convolution of iid exponential random variables c) Minimum of an independent family of exponential random variables

#4
Front (Question)

Exponential distribution ▪ Properties a) Loss of memory

Back (Answer)

Consider ▪ X ~ Exp(λ) ▪ It s,t>0 P ( X>s+t|X>t ) = P ( X>s ) P ( X>µ ) = 1 - Fx(µ) = e^(-λµ)

#5
Front (Question)

Exponential distribution ▪ Properties b) Convolution of iid exponential random variables

Back (Answer)

Let X₁, …, Xₙ are iid exponential random variables with parameter λ. Then X₁, …, Xₙ has gamma distribution with parameters n and λ X₁, …, Xₙ iid~ Exp(λ) → X₁, …, Xₙ~ Γ(n, λ)

#6
Front (Question)

Exponential distribution ▪ Gamma distribution

Back (Answer)

A continuous random variable X is said to have Gamma distribution with parameter α,λ>0 X ~ Γ(α, λ) Density: fₓ(x)= ▪ (λᵃ/Γ(α))xᵃ⁻¹e^(-λx) if x⩾0 ▪ 0 if x<0 where Γ(α)=₀∫⁺∞ xᵃ⁻¹*e^(-x) dx Then E(X) = α/λ Var(X) = α/λ²

#7
Front (Question)

Exponential distribution ▪ Properties c) Minimum of an independent family of exponential random variables

Back (Answer)

Suppose that X₁, …, Xₙ are independent exponential random variables with: Xᵢ ~ Exp(λᵢ) –> (i = 1, …, n) The smallest of the Xᵢ has exponential distribution with parameter the sum of the Xᵢ Proposition If are independent with: Xᵢ ~ Exp(λᵢ) –> (i = 1, …, n) Then ▪ min{X₁, …, Xₙ}~ Exp(λ₁, …, λₙ) ▪ P(Xj = min{X₁, …, Xₙ})=λj/(λ₁+…+λₙ)