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3. Gamma distribution

3. Gamma distribution. ∞ β -1 -x 0. For all β > 0 Γ ( β ) = ∫ x e dx (5.30) Γ ( β ) = ( β - 1) Γ ( β -1) (5.31) If β is an integer, then using Γ (1) = 1, Γ ( β ) = ( β - 1)! (5.32) f(x) = ( βθ x) e ,x>0 (5.33)

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3. Gamma distribution

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  1. 3. Gamma distribution

  2. ∞ β-1 -x 0 • For all β > 0 Γ(β) =∫ x e dx (5.30) • Γ(β) = (β - 1)Γ(β-1) (5.31) • If β is an integer, then using Γ(1) = 1, Γ(β) = (β - 1)! (5.32) • f(x) = (βθx) e ,x>0 (5.33) 0, otherwise { βθβ-1 - βθx Γ(β)

  3. 1 θ 1 βθ • E(X) = (5.34) • V(X) = (5.35) • F(x) = 1 - ∫(βθt) e dt, x>0 0, x<0 (5.36) • X = X + X + … + X (5.37) • g(x) = (βθ)e , x>0 0, otherwise 2 ∞ βθβ-1 -βθt x Γ(β) { 1 2 β { -βθxj

  4. Erlang distribution

  5. Erlang distribution • Erlang was a Danish telephone engineer who was an early developer of queueing theory. • About Erlang distribution： • 假設有連續k個站，而顧客必須通過這些站來完成服務。 • 一個新的顧客不能進入第一個站，必須等到正在服務中的顧客順利通過所有的站。 • 每一站的服務時間是依一個參數kθ呈指數分佈。

  6. Erlang distribution • E(X) = 1/θ • V(X) = 1/βθ2 • 根據β的值決定gamma的中間值以及變動程度 • 然而當β = k (k為一整數)時， X = X1 + X2 + ‧‧‧+ Xk

  7. Erlang distribution • E(X) = E(X1) + E(X2) + ‧‧‧+E(Xk) • E(X) = 1/kθ + 1/kθ+ ‧‧‧+ 1/kθ=1/θ • V(X) = 1/(kθ)2 + 1/(kθ)2 + ‧‧‧+ 1/(kθ)2 =1/kθ 2 • k為一整數的情況下5.36的方程式便會變成5.38

  8. EXAMPLE 5.19 • 一個大學教授整個暑假不在家，為了防止竊賊而必須整天開燈。 • 他裝了一個裝置可以切換兩個燈泡以防止第一個燈泡故障。 • 平均一個燈泡可以用1000小時。 • 教授有90天(2160小時)不在家。 • 教授回來時燈炮還亮著的機率？

  9. EXAMPLE 5.19(續) • R(X) = 1 – F(X) (亮 = 1- 不亮) • β = k = 2 (2個燈泡) • kθ=1/1000 →θ=1/2000 • X = 2160 • F(2160) = 0.636 → R(X) = 36%

  10. EXAMPLE 5.20 • 一個病患在做健康檢查時必須通過3個檢驗站。 • 每個站平均要花20分鐘。 • 在50分鐘內完成檢查的機率是多少？

  11. EXAMPLE 5.20(續) • k = 3 • kθ = 1/20 → θ = 1/60 • F(50) = 0.457 = 45.7% • E(X) = 1/θ = 60min • V(X) = 1/kθ2 = 1200min2 • mode = k-1/kθ(Erlang distribution) • mode = 3-1/3(1/60) = 40min

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