ECE 534: Random Pro cesses Spring 2010
Solutions to Problem Set 6
3.21 Estimation for an additive Gaussian noise model
The vectors xand yare jointly Gaussian since they are obtained from xand nvia linear
operations. Thus, the conditional distribution of xgiven yis a Gaussian distribution. For
a given value of y, that distribution has mean ¯x+ ΣxyΣ−1
yy (y−¯y) and covariance matrix
equal to Σee = Σxx −ΣxyΣ−1
yy Σyx where ¯x=Ex, ¯y=Ey, ¯y=E[x+n] = ¯x+ ¯n,
Σxy = Cov(x, x +n)=Σx+ Σxn = Σx,and
Σyy = Cov(x+n, x +n)=Σx+ Σn.
The proof of the equivalent formulas is left to the reader.
3.23 Steady state gains for one-dimensional Kalman filter
(a) Let bk=σ2
k. The solution to Problem 3.22 shows that the sequence bksatisfies the
recursion bk+1 =F(bk), where F(b) = bf 2
1+b+ 1, with the initial condition b0=σ2≥0
given. The function Fis positive, strictly increasing, and bounded. Thus if bk≤bk+1
for some k, then bk+1 =F(bk)≤F(bk+1) = bk+2 . Therefore, if b0≤b1, then the
sequence (bk:k≥0) is monotone nondecreasing and bounded. Simlarly, if b0≥b1
then the sequence (bk:k≥0) is monotone nonincreasing and bounded. Since bounded
monotone sequences have finite limits, the sequence (bk:k≥0) converges.
(b) Denote the limit by b∞(so b∞=σ2
∞). Since Fis a continuous function, letting k→ ∞
in the equation bk+1 =F(bk) yields b∞=F(b∞), which has the unique nonnegative
solution b∞=f2+√f4+4
2.
(c) If f= 0, the states xkare uncorrelated with variance one. The observations y0, . . . , yk−1
are therefore orthogonal to xk, and the variance of the error, σ2
k, is just the variance of
xk, equal to one for all k. The limiting variance of error is thus also one.
3.25 The Kalman filter for xk|k
By time kwe have already calculated bxk|k−1. In fact, it was already done at time k−1.
The problem is to update bxk|k−1by incorporating the new observation yk, or equivalently
the new observation eyk. Therefore,
bxk|k=bxk|k−1+Ak
eyk
where eyk=yk−HT
k
bxk|k−1and
Ak= Cov(xk,eyk)Cov(eyk,eyk)−1
= Cov(xk−bxk|k−1, HT
k(xk−bxk|k−1) + vk)Cov(eyk,eyk)−1
= Σk|k−1Hk[HT
kΣk|k−1Hk+Rk]−1.
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