Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Lecture 10: Other Continuous Distributions and ProbabilityPlotsDevore: Section 4.4-4.6October, 2011Page 1Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Gamma Distribution• Gamma function is a natural extension of the factorial• For any α > 0,Γ(α) =Z∞0xα−1e−xdx• Properties:1. If α > 1, Γ(α) = (α − 1)Γ(α − 1)2. Γ(n) = (n − 1)! for any n ∈ Z+3. Γ12=√πOctober, 2011Page 2Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011• The natural definition of a density based on the gamma functionisf(x; α) =xα−1e−xΓ(α)if x ≥ 00 otherwise• A gamma density with parameters α > 0, β > 0 isf(x; α, β) =1βαΓ(α)xα−1e−x/βif x ≥ 00 otherwise• α is a shape parameter, β is a scale parameter• The case β = 1 is called the standard gamma distributionOctober, 2011Page 3Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Gamma pdf: graphical illustrationOctober, 2011Page 4Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Gamma Distribution Parameters• The mean and variance of a random variable X having thegamma distribution f(x; α, β) areE(X) = αβ and V (X) = αβ2• Let X have a gamma distribution with parameters α and β• Then P (X ≤ x) = F (x; α, β) = F (x/β; α)• In the above,F (x; α) =Zx0yα−1e−yΓ(α)dyis an incomplete gamma function. It is defined for any x > 0.October, 2011Page 5Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Exponential Distribution as a Special Case of Gamma Distribution• Assume that α = 1 and β =1λ.• Then,f(x; λ) =λe−λxif x ≥ 00 otherwise• Its mean and variance areE(X) =1λand V (X) =1λ2• Note that µ = σ in this caseOctober, 2011Page 6Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Exponential pdf: graphical illustrationOctober, 2011Page 7Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Exponential cdf• Exponential cdf can be easily obtained by integrating pdf, unlikethe cdf of the general Gamma distribution• The result isF (x; λ) =1 − e−λxif x ≥ 00 otherwiseOctober, 2011Page 8Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Example I• Suppose the response time X at an on-line computer terminalhas an exponential distribution with expected response time 5sec• E(X) =1λ= 5 and λ = 0.2• The probability that the response time is between 5 sec and 10sec isP (5 ≤ X ≤ 10) = F (10; 0.2) − F (5; 0.2) = 0.233• In R: pexp(10,0.2)-pexp(5,0.2)October, 2011Page 9Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Example II• On average, 3 trucks per hour arrive to a given warehouse to beunloaded. What is the probability that the time between arrivalsis less than 5 min? At least 45 min?• Arrivals follow a Poisson process with parameter λ = 3; therespective exponential distribution parameter is also λ = 3•Z1/1203 exp(−3x) dx = 1 − exp(−1/4) = 0.221•Z∞3/43 exp(−3x) = exp(−9/4) = 0.105October, 2011Page 10Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Weibull Distribution• X is said to have a Weibull distribution iff(x; α, β) =αβαxα−1e−(x/β)αif x ≥ 00 otherwise• Note that Weibull distribution is yet another generalization ofexponential; indeed, if α = 1, Weibull pdf is exponential withλ =1β.• The Weibull cdf isF (x; α, β) =1 − e−(x/β)αif x ≥ 00 otherwiseOctober, 2011Page 11Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011• The parameters α > 0 and β > 0 are referred to as the shapeand scale parameters, respectively.October, 2011Page 12Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Weibull densities: graphical illustrationOctober, 2011Page 13Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Alternative definition of Weibull distribution• Weibull distribution is used in survival analysis, many processesin engineering, such as engine emissions, degradation dataanalysis etc• Often, the Weibull distribution is also defined as bounded frombelow.• Let X be the corrosion weight loss for a small squaremagnesium alloy plate. The plate has been immersed for 7 daysin an inhibited 20% solution of MgBr2.• Suppose the min possible weight loss is γ = 3 and X = 3 hasa Weibull distribution with α = 2 and β = 4.October, 2011Page 14Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011•F (x; 2, 4, 3) =1 − e−[(x−3)/4]2if x ≥ 30 if x < 3• Then, e.g. probabilityP (X > 3.5) = 1 − F (3.5; 2, 4, 3) = .985October, 2011Page 15Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Lognormal distribution• If Y = log(X) is normal, X is said to have a lognormaldistribution• Its pdf is1√2πσxe−[log(x)−µ]2/(2σ2)if x ≥ 00 if x < 0October, 2011Page 16Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Lognormal densities: graphical illustrationOctober, 2011Page 17Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011• Note that µ and σ2are NOT the mean and variance of thelognormal distribution. Those are E(X) = exp(µ + σ2/2)and V ar(X) = exp(2µ + σ2)(exp(σ2) − 1). Also note thatthe lognormal distribution is not symmetric but rather positivelyskewed.• Lognormal distribution is commonly used to model variousmaterial properties.• To find probabilities related to the lognormal distribution notethatP (X ≤ x) = P (log(X) ≤ log(x))= PZ ≤log(x) − µσ= Φlog(x) − µσOctober, 2011Page 18Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Beta Distribution• Beta Distribution is used to model random quantities that arepositive on a closed interval [A, B] only. It gives us much moreflexibility than the uniform distribution.• A RV X is said to have a beta distribution with parameters A,B, α > 0 and β > 0 if its pdf isf(x; α, β, A, B) =1B−A·Γ(α+β)Γ(α)Γ(β)x−AB−Aα−1B−xB−Aβ−1if A ≤ x ≤ B0 if x < 0• The mean and variance of X are µ = A + (B − A)αα+βandσ2=(B−A)2αβ(α+β)2(α+β+1).October, 2011Page 19Statistics 511: Statistical MethodsDr. LevinePurdue UniversityFall 2011Beta densities: