# Reply to Reviewer B Many Thanks for your valuable comments and

```Reply to Reviewer B
Many Thanks for your valuable comments and suggestions which have led to significant improvement on
the presentation and quality of this paper. In what follows, we shall detail the changes we have made on
the paper.
1. In the abstract, the authors mention “an analytical solution to the filter is obtained”. I think this
may confuse readers who will expect to find a closed-form analytic solution from the paper.
• Reply: Following your suggestions, the words “And an analytical solution to the filter is obtained by
solving the Riccati difference equations” is replaced by “And the solution is given in terms of two
Riccati difference equations”.
2. On page 2, different R is used for defining diag{Dj }.
• Reply: Thanks very much for your carefulness, we have changed R as R throughout the paper.
3. There are some undefined notations: col, cov (on page 3 and others).
• Reply: We have added the definitions of “ col, cov” in the last paragraph of the Introduction.
4. It is rather confusing to use the gradient operator to denote the end of a proof.
• Reply: I have deleted the gradient operators in all proofs.
5. On page 8, first line in the proof, “r.v.” should be “random variables”.
• Reply: Following your comments, the words “r.v.” have been changed as “random variables”.
6. In the formulation, if the instantaneous measurements (2.2) cannot be obtained completely, will the
solution procedure remain to be the same?
• Reply: If there exist dropouts in (2.2), the estimator design procedure remain to the same as that
proposed in this paper. We outline the concerning result as follows: Consider the following system
x(k + 1) = Aθ(k) x(k) + Cθ(k) w(k),
y(k) = γk Hθ(k) x(k) + Gθ(k) v(k),
y1 (k) =
H̃θ(k) x(k − d) + G̃θ(k) v1 (k),
where the arrival of the measurement is described by a binary random variable γk , which takes values
of 0 or 1 with probability distribution
q = P r{γk = 1}, 1 − q = P r{γk = 0}, E(γk ) = E(γk2 ) = q.
Follow the same steps as in the paper, we will obtain the linear minimum mean square error estimator
x̂(k|k) =
N
X
ẑi (k, 1|k),
i=1
where the estimator ẑi (k, 1|k) (ẑi (k, 1|k) is the ith element of the vector ẑ(k, 1|k)) is computed
following the steps below:
• For 0 ≤ s ≤ k1 , ẑ(s, 2|s) can be calculated by the following difference equation with the initial value
of ẑ(0, 2| − 1) = 0,
ẑ(s, 2|s) = ẑ(s, 2|s − 1) + K2 (s)ε2 (s),
ẑ(s, 2|s − 1) = Āẑ(s − 1, 2|s − 1),
where
ε2 (s) =
K2 (s) =
diag{qI, I}H̄ z̃(s, 2) + diag{(γs − q)I, I}H̄z(s, 2) + Ḡθ(s) v̄(s),
P (s, 2|s − 1)H̄ T diag{qI, I}(diag{qI, I}H̄P (s, 2|s − 1)H̄ T diag{qI, I}
+diag{q(1 − q)I, I}H̄Z(s, 2)H̄ T + Ḡ(s)ḠT (s))−1 ,
and P (s + 1, 2|s) satisfies the following Riccati equation:
N X
N
X
ĀP (s, 2|s − 1)Ā + diag{
πl (s)πm (s + d)λli λmj Cl Cl0 }
0
P (s + 1, 2|s) =
l=1 m=1
P (0, 2| − 1)
−ĀK2 (s)diag{qI, I}H̄P (s, 2|s − 1)Ā0 + B(Z(s, 2)),
= diag{πi (0)πj (d)V }.
where Ā, B(Z(s, 2)) and Z(s, 2) remain the same as (3.24), (3.26) and (3.43) in the paper.
• For k1 &lt; s ≤ k, ẑ(s, 1|s) can be computed by the following recursive equation
ẑ(s, 1|s) =
ẑ(s, 1|s − 1) =
ẑ(s, 1|s − 1) + K1 (s)ε1 (s),
Aẑ(s − 1, 1|s − 1),
where
ε1 (s) =
K1 (s) =
P (s + 1, 1|s) =
qH z̃(s, 1|s − 1) + (γs − q)Hz(s, 1) + Gθ(s) v(s),
qP (s, 1|s − 1)H 0 [q 2 HP (s, 1|s − 1)H T + q(1 − q)HZ(s, 1)H 0
+G(s)G(s)0 ]−1 ,
AP (s, 1|s − 1)AT + B(Z(s, 1)) + diag{
N
X
πl (s)λli Cl Cl0 }
l=1
T
−qAK1 (s)HP (s, 1|s − 1)A .
P (k1 + 1, 1|k1 ) =
N
X
diag{
Pij (k1 + 1, 2|k1 )},
j=1
where A, B(Z(s, 1)) and Z(s, 1) remain the same as (3.28), (3.30), and (3.48) in the paper. It can
be seen that the differences between the above result and that proposed in the paper, exist in the
Riccati difference equations and the innovation processes.
7. In the numerical solution, a simple problem about maneuvering target tracking via a wireless sensor
network is presented. I think the authors can elaborate and explain the setup more. For example, what is
the interpretation of the markov process θ(k) and what might be the delay d here? after defining (4.55),
what will be the Riccati difference equations look like? How to obtain the true position of the trajectory?
• Reply: We have improved the presentation of the simulation to make the problem setup more clear,
and added the interpretation of θ(k) (see the last two sentences on page 12) and delay d (see the
context before (4.55) on page 13). When the parameters are given, the Riccati equation can be
obtained according to Theorem 3.1. In this simulation, the system is with dimension 4 &times; 1, then the
Riccati equation’s dimension in (3.23) is 16 &times; 16, and the Riccati equation’s dimension in (3.27) is
8 &times; 8. So we think that section 4 is not concise if the details
example, the parameters Ā and H̄ in (3.23) are as follows.

0.7225 0.595
0
0
0.0723 0.0595
 0.1275 0.255
0
0
0.0128 0.0255


0
0
0.7225 0.595
0
0


0
0
0.1275 0.255
0
0


0
0
0
0
0.7225 0.595


0
0
0
0
0.1275 0.255


0
0
0
0
0
0


0
0
0
0
0
0
Ā = 
 0.1275 0.105
0
0
0.0128 0.0105

 0.0225 0.045
0
0
0.0023
0.0045


0
0
0.1275
0.105
0
0


0
0
0.0225
0.045
0
0


0
0
0
0
0.1275
0.105


0
0
0
0
0.0225 0.045


0
0
0
0
0
0
0
0
0
0
0
0

1
 0
H̄ = 
 1
0
0
1
0
1
0
0
0
0
0
0
0
0
1
0
1
0
0
1
0
1
of the Riccati equations are added. For
0
0
0.595
0.105
0
0
0
0
0
0
0.255
0.045
0
0
0
0
0
0
0.49
0.21
0
0
0
0
0
0
0.21
0.09
0
0
0
0
0
0
0.595 0.49
0
0
0.105 0.21
0.0723 0.0595
0
0
0.0128 0.0255
0
0
0
0
0
0
0
0
0
0
0.7225 0.595
0
0
0.1275 0.255
0
0
0
0
0.255 0.21
0
0
0.045 0.09
0.0128 0.0105
0
0
0.0023 0.0045
0
0
0
0
0
0
0
0
0
0
0.1275 0.105
0
0
0.0225 0.045
0
0

0.0595 0.049
0
0
0.0101 0.021
0
0 

0
0
0.0595 0.049 

0
0
0.0105 0.021 

0.595
0.49
0
0 

0.105
0.21
0
0 

0
0
0.595
0.49 

0
0
0.105
0.21 
.
0.0255 0.021
0
0 

0.0045 0.009
0
0 

0
0
0.0255 0.021 

0
0
0.0045 0.009 

0.255 0.21
0
0 

0.045 0.09
0
0 

0
0
0.255
0.21 
0
0
0.045
0.09
0
0
0
0
1
0
1
0
0
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
1
0
1
0
0
1
0
1
0
0
0
0

0
0 
.
0 
0
In practice, the true trajectory of the target is difficult or impossible to obtain. It can only be obtained
under lab environment. In this simulation, in order to illustrate the performance of the proposed estimator,
we assume that the true trajectory of the target is exactly known.
```