(∂) = pdf of Gamma distribution, f(x): 1 = ∫ 1 x α−1 e−xdx, f (x) = { 1 x ...

�(∂) =

�

∗

x∂−1 e−x dx

0

p.d.f of Gamma distribution, f(x): � ∗ 1 ∂−1 −x 1 ∂−1 −x x e dx, f (x) = { x e , x ∼ 0; 0, x < 0} 1= �(∂) �(∂) 0 Change of variable x = λy, to stretch the function: � ∗ ∂ � ∗ 1 ∂−1 ∂−1 −ξy λ 1= λ y e λdy = y ∂−1 e−ξy dy �(∂) �(∂) 0 0

p.d.f. of Gamma distribution, f (x|∂, λ): f (x|∂, λ) = {

λ ∂ ∂−1 −ξx x e , x ∼ 0; 0, x < 0} − Gamma(∂, λ) �(∂)

Properties of the Gamma Function: � �(∂) =

∗

x∂−1 e−x dx =

0

Integrate by parts:

= x∂−1 e−x |∗ 0 −

�

∗ 0

�

∗

x∂−1 d(−e−x ) =

0

(−e−x )(∂ − 1)x∂−2 dx = 0 + (∂ − 1)

�

∗ 0

x∂−2 e−x dx = (∂ − 1)�(∂ − 1)

In summary, Property 1: �(∂) = (∂ − 1)�(∂ − 1) You can expand Property 1 as follows: �(n) = (n − 1)�(n − 1) = (n − 1)(n − 2)�(n − 2) = (n − 1)(n − 2)(n − 3)�(n − 3) = = (n − 1)...(1)�(1) = (n − 1)!�(1), �(1) = In summary, Property 2: �(n) = (n − 1)!

�

∗ 0

e−x dx = 1 ↔ �(n) = (n − 1)!

Moments of the Gamma Distribution: X ≈ (∂, λ) � ∗ � ∗ ∂ λ∂ k k λ ∂−1 −ξx EX = x x e dx = x(∂+k)−1 e−ξx dx �(∂) �(∂) 0 0

Make this integral into a density to simplify: � λ ∂ �(∂ + k) ∗ λ ∂+k x(∂+k)−1 e−ξx dx = �(∂) λ ∂+k �(∂ + k) 0 The integral is just the Gamma distribution with parameters (∂ + k, λ)! =

�(∂ + k) (∂ + k − 1)(∂ + k − 2) × ... × ∂�(∂) (∂ + k − 1) × ... × ∂ = = �(∂)λ k �(∂)λ k λk

For k = 1:

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