Probabilistic Performance Analysis of Fault Diagnosis Schemes

Proof. Using the error function, defined in Section 2.2.6, we can write the function L as
L(ν,µ) = 1 2
[ ( ) ( ν − µ ν + µ
erf + erf
)].
2 2
Since the map µ → L(ν,µ) is clearly even, it suffices to consider 0 ≤ µ 1 < µ 2 . We prove the
claim by showing that µ → L(ν,µ) is monotonically decreasing on [0,∞). The derivative of L
at µ 0 ≥ 0 is
∂L(ν,µ)
∂µ
∣ = 1 ∂
µ=µ0 2 ∂µ
= 1 2
[ ( ) ( )]
ν − µ ν + µ
erf + erf
2 2
[ 2 π exp
= 1
2π
[exp
(− (ν − µ 0) 2
2
(− (ν + µ 0) 2
2
µ=µ 0
)( ) −1
+ 2 exp
(− (ν + µ 0) 2
2 π
)
− exp
(− (ν − µ 0) 2
2
)]
.
2
)( 12
)]
Since µ 0 ≥ 0,
(ν − µ 0 ) 2 ≤ (ν + µ 0 ) 2 ,
with equality if and only if µ 0 = 0. This inequality, together with the fact that the map
x → e −x is monotonically decreasing, implies that
with equality if and only if µ 0 = 0.
∂L(ν,µ)
∂µ
∣ ≤ 0,
µ=µ0
Proof of Proposition 5.2. Define the “scaled” residual
µ k (ρ) := r k(ρ)
Σk
,
and let ν > 0 be such that ε k = νΣ k , for all k. Note that the conditional mean of µ k (ρ) is
ˆµ k (ρ,ϑ 0:k ) := E ( µ k (ρ) | θ 0:k = ϑ 0:k
)
= ˆr k (ρ,ϑ 0:k )
Σk
,
and the conditional variance of µ k (ρ) is
( (µk
E (ρ) − ˆµ k (ρ,ϑ 0:k ) ) 2 ∣ )
θ0:k = ϑ 0:k = 1 ( (rk
E (ρ) − ˆr k (ρ,ϑ 0:k ) ) 2 ∣ )
θ0:k = ϑ 0:k = 1.
Σ k
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