A proof of an equation containing improper gamma distribution
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This document proves the equation exp(-∞∫a(1 - e-tz)z-1e-zdz) = (1 + t)-a by taking the logarithm of both sides and rewriting the integral using Taylor series expansions of the exponential and logarithm functions. The integral is expanded as the sum from 0 to infinity of tn/n!Γ(n), which is shown to equal the Taylor series expansion of ln(1 + t), proving the original equation.
A proof of an equation containing improper gamma distribution
- 1. A proof of an equation containing improper gamma distribution Tomonari MASADA @ Nagasaki University January 21, 2013 We’d like to prove the following equation: exp − ∞ 0 a(1 − e−tz )z−1 e−z dz = (1 + t)−a . (1) This equation appears on page 92 of Poisson Processes by J. F. C. Kingman, where we replace µ(A) by a. By taking the logarithm of both sides, we obtain ∞ 0 (1 − e−tz )z−1 e−z dz = ln(1 + t) (2) and would like to prove this. e−tz can be expanded as follows: e−tz = 1 − t 1! z + t2 2! z2 − t3 3! z3 + t4 4! z4 − · · · . (3) Therefore, we can rewrite the integral ∞ 0 (1 − e−tz )z−1 e−z dz as follows: ∞ 0 (1 − e−tz )z−1 e−z dz = ∞ 0 t 1! z − t2 2! z2 + t3 3! z3 − t4 4! z4 + · · · z−1 e−z dz = t 1! ∞ 0 z1−1 e−z dz − t2 2! ∞ 0 z2−1 e−z dz + t3 3! ∞ 0 z3−1 e−z dz − t4 4! ∞ 0 z4−1 e−z dz + · · · = t 1! Γ(1) − t2 2! Γ(2) + t3 3! Γ(3) − t4 4! Γ(4) + · · · = t 1 − t2 2 + t3 3 − t4 4 + · · · , (4) where we use the gamma integral formula, i.e., ∞ 0 zb−1 e−az dz = Γ(b) ab . On the other hand, ln(1 + t) can be expanded as follows: ln(1 + t) = 0 + 1 1 + t 0 · t 1! − 1! (1 + t)2 0 · t2 2! + 2! (1 + t)3 0 · t3 3! − 3! (1 + t)4 0 · t4 4! + · · · = t 1 − t2 2 + t3 3 − t4 4 + · · · . (5) Therefore, ∞ 0 (1 − e−tz )z−1 e−z dz = ln(1 + t) . (6) 1