A.11 Proof of Proposition 8
As shown before, if $g(\theta_i, e_i) = \theta_i e_i$, we have
( d d e i g e k − 1 ( g e i ( t i + x ) ) ) ∣ e 1 = ⋯ = e n = e = ( t i + x ) e . ( d e i d g e k − 1 ( g e i ( t i + x )) ) e 1 = ⋯ = e n = e = e ( t i + x ) .
Thus, making use of expression (11), derived in Section B.1, and denoting $\Delta t = t_1 - t > 0$,
λ 2 ∫ − Δ t ( H ( x ) ∏ k ≠ i H ( Δ t + x ) ) ( ∑ k ≠ i ( d d e i g e k − 1 ( g e i ( t 1 + x ) ) ) ) ∣ e 1 = ⋯ = e n = e d x = λ 2 ∫ − Δ t exp ( n λ y ) ⋅ exp ( ( n − 1 ) λ ( Δ t + x ) ) ( n − 1 ) ( t 1 + x ) e d x = λ 2 ( n − 1 ) e ∫ − Δ t exp ( n λ y + ( n − 1 ) λ Δ t ) ( t 1 + y ) d y . λ 2 ∫ − Δ t H ( x ) k = i ∏ H ( Δ t + x ) k = i ∑ ( d e i d g e k − 1 ( g e i ( t 1 + x )) ) e 1 = ⋯ = e n = e d x = λ 2 ∫ − Δ t exp ( nλ y ) ⋅ exp (( n − 1 ) λ ( Δ t + x )) ( n − 1 ) e ( t 1 + x ) d x = e λ 2 ( n − 1 ) ∫ − Δ t exp ( nλ y + ( n − 1 ) λ Δ t ) ( t 1 + y ) d y .
The map $\phi_2 : \mathbb{R}_x \to \mathbb{R}_y$ given by $x \to y = -\Delta t + x$ is a smooth diffeomorphism with $\det|\phi_2'(x)| = 1$. Applying the associated change of variables to the integral, we obtain
λ 2 ( n − 1 ) e ∫ 0 exp ( n λ ( x − Δ t ) + ( n − 1 ) λ Δ t ) ( t + x ) d x = ( n − 1 ) e exp ( − λ Δ t ) λ 2 ( ∫ 0 x exp ( n λ x ) d x + t ∫ 0 exp ( n λ x ) d x ) . e λ 2 ( n − 1 ) ∫ 0 exp ( nλ ( x − Δ t ) + ( n − 1 ) λ Δ t ) ( t + x ) d x = e ( n − 1 ) exp ( − λ Δ t ) λ 2 ( ∫ 0 x exp ( nλ x ) d x + t ∫ 0 exp ( nλ x ) d x ) .
Notice that
n λ ∫ 0 x exp ( n λ x ) d x nλ ∫ 0 x exp ( nλ x ) d x
is the mean of a random variable that is distributed according to the reflected exponential distribution with parameter $n\lambda$, hence
n λ ∫ 0 x exp ( n λ x ) d x = − 1 n λ nλ ∫ 0 x exp ( nλ x ) d x = − nλ 1
⇔ ∫ 0 x exp ( n λ x ) d x = − 1 n 2 λ 2 . ⇔ ∫ 0 x exp ( nλ x ) d x = − n 2 λ 2 1 .
Furthermore,
n λ ∫ 0 exp ( n λ x ) d x = 1 nλ ∫ 0 exp ( nλ x ) d x = 1
⇔ ∫ 0 exp ( n λ x ) d x = 1 n λ . ⇔ ∫ 0 exp ( nλ x ) d x = nλ 1 .
