Chapter V: Selected Solutions
H. V. PoorPrinceton University
October 14, 2002
Exercise 3:
Since sk depends completely on Yk0, it behaves like a constant
when conditional quantities
given Y k0 are computed. The only point at which this aects the
derivation of the Kalman-Bucy lter is in the time update equations,
which now become:
X t+1|t = FtX t|t + E{tst|Y t0} = FtX t|t + tst,and
t+1|t = Ftt|tFTt + GtQtGTt + Cov
(tst|Y t0
)= Ftt|tFTt + GtQtG
Tt .
Note that the second of these two equations is the same as when
there is no measurementfeedback.
The measurement feedback has no eect on the measurement update
equations. Al-though the presence of this feedback may cause the
state to be nonGaussian, the jointconditional statistics of X t and
Y t given Y
t10 are still Gaussian. Thus, the measurement
update is unchanged from the case of no measurement feedback
since it depends only onthis joint Gaussian property and the
linearity of measurement equation.
Exercise 4:
There are several ways of approaching this problem. An
interesting one is to note that,although U t and V t are dependent,
the Gaussian vector U
t U tCtR1t V t is independent
of V t. To take advantage of this, we may add the zero quantity
CtR1t [Y t HtX t V t]
to the tth state input, which yields the equivalent state
equation
X t+1 = FtX t + GtUt + GtCtR
1t (Y t HtX t) .
So, we have an equivalent problem with independent state and
measurement noises,but with the measurement feedback term GtCtR
1t Y t, and with the new state matrix(
Ft GtCtR1t Ht). We also have a dierent correlation matrix for
the state input, since
Cov(U t) = Qt CtR1t CTt .
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Applying the result of Exercise 3 and eliminating the
measurement update equationsyields the given result.
Exercise 6:
a. This result follows by induction on t. We rst note that, for
t = j, the given equalityfollows by denition. From the state
equation we have that, for k > j,
E{(Xj Xj|k
)XTk+1
}= E
{(Xj Xj|k
)(FkXk + GkUk)
T}
= E{(Xj Xj|k
)XTk
}FTk ,
where we use the fact that Uk has zero mean and is independent
of both Xj and Xj|k.Now, on applying the recursion for Xj|k to this
equation we have
E{(Xj Xj|k
)XTk+1
}= E
{(Xj Xj|k1
)XTk
}FTk KakE
{(Y k HkXk|k1
)XTk
}FTk .
We now assume that the given equality is true for t = k. From
this and the denition ofKak, we then have
E{(Xj Xj|k
)XTk+1
}= ak|k1
[IHTk
(Hkk|k1HTk + Rk
)1E
{(Y k HkXk|k1
)XTk
}]FTk
= ak|k1
[IHTk
(Hkk|k1HTk + Rk
)1HkE
{(Xk Xk|k1
)XTk
}]FTk
= ak|k1
[IHTk
(Hkk|k1HTk + Rk
)1Hkk|k1
]FTk =
ak|k1
[IHTkKTk
]FTk ,
where the last equality follows by deniton of Kk. Applying the
recursion for ak+1|k, we
haveE
{(Xj Xj|k
)XTk+1
}= ak+1|k,
which shows that the given equation for t = k+ 1. The induction
principle thus gives thedesired result.
b. As with the measurement-update derivation in the Kalman-Bucy
lter, we notehere that Xj and Y t are jointly Gaussian conditioned
on Y
t10 . Thus, E
{Xj|Y t0
}will be
as given by the recursion if we have
Cov(Xj, Y t|Y t10
) [Cov
(Y t|Y t10
)]1= Kat .
But, since Y t = HtX t + V t, the result from Part a. implies
this equality.
Exercise 9:
This is the Kalman-Bucy problem with all dimensions equal to
unity, Fk 1,Gk Qk 0,Hk = sk,Rk 2,m0 ,0 v2, Xn+1|n Xn|n n, and n|n
n|n E
{(n
)2}. The desired recursions thus follow by eliminating either
set of updates
from (V.B.14) - (V.B.16). The resulting estimate is the same as
that found in ExampleIV.B.2.
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