Probabilistic Performance Analysis of Fault Diagnosis Schemes

Fact 4.1. Given a Markov chain θ ∼ ( Θ,{Π k },π 0
)
, let l > 0 and ϑ0:l ∈ Θ l+1 . If Π k (i , j ) can be
computed or retrieved in O(1) time, for any k ≥ 0 and any i , j ∈ Θ, then
can be computed in O(l) time.
p θ (ϑ 0:l ) = P(θ 0:l = ϑ 0:l )
Proof. By definition, P(θ 0 = ϑ 0 ) = π 0 (ϑ 0 ). Let 0 < τ ≤ l. Because {θ k } is Markov, the probability
of the event {θ 0:τ = ϑ 0:τ } can be factored as
Hence, by induction on τ,
P(θ 0:τ = ϑ 0:τ ) = P(θ τ = ϑ τ | θ 0:τ−1 = ϑ 0:τ−1 ) P(θ 0:τ−1 = ϑ 0:τ−1 )
= P(θ τ = ϑ τ | θ τ−1 = ϑ τ−1 ) P(θ 0:τ−1 = ϑ 0:τ−1 )
= Π τ−1 (ϑ τ−1 ,ϑ τ ) P(θ 0:τ−1 = ϑ 0:τ−1 ).
P(θ 0:l = ϑ 0:l ) = Π l−1 (ϑ l−1 ,ϑ l ) Π l−2 (ϑ l−2 ,ϑ l−1 )···Π 0 (ϑ 0 ,ϑ 1 ) π 0 (ϑ 0 ).
Since this computation requires l evaluations of the transition probability matrices and l
scalar multiplications, the overall time-complexity is lO(1) + lO(1) = O(l).
The second computational issue raised in Section 4.1 is the high dimensionality of
the integral in equation (4.1). Since the fault parameter space Θ is assumed to be finite,
equation (4.1) can be written as
J k (i , j ) =
∑
ϑ 0:k ∈Θ k ×Θ i
P(D j,k | θ 0:k = ϑ 0:k ) P(θ 0:k = ϑ 0:k ), (4.3)
for all i , j ∈ D and all k ≥ 0. Of course, exchanging an integral for a summation is of little
use if the summation has an intractable number of terms (i.e., the number of terms grows
exponentially with k). In general, the summation (4.3) has m k m i terms, where m i := |Θ i |.
The following example illustrates the practical implications of this exponential growth.
Example 4.2. Suppose that {y k } k≥0 is a stochastic process taking values in R such that the conditional
density p y|θ (y k | θ 0:k = ϑ 0:k ) is Gaussian for all k and all ϑ 0:k ∈ Θ k+1 . Then, the marginal
density of y k can be written as the sum
p y (y k ) =
∑
p y|θ (y k | θ 0:k = ϑ 0:k ) P(θ 0:k = ϑ 0:k ).
ϑ 0:k ∈Θ k+1
In this sum, each term is represented by three scalars: the mean and variance of the Gaussian
density p y|θ (y k | θ 0:k = ϑ 0:k ) and the probability P(θ 0:k = ϑ 0:k ). If these data are stored in ieee
single precision (i.e., 32 bits per number), then each term requires 3 × 32 = 96bits or 12bytes
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