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University of Illinois at Urbana-Champaign
ECE 434: Random Processes
Spring 2003 Exam 1
Monday, March 10, 2003
Name:
- You have 75 minutes for this exam. The exam is closed book and closed note, except that you may consult both sides of one sheet of notes, typed in font size 10 or equivalent handwriting size.
- Calculators, laptop computers, Palm Pilots, two-way e-mail pagers, etc. may not be used.
- Write your answers in the spaces provided.
- Please show all of your work. Answers without appropriate justification will receive very little credit. If you need extra space, use the back of the previous page.
Score:
- (12 pts.)
- (12 pts.)
- (8 pts.)
- (8 pts.)
Total: (40 pts.)
Problem 1 (12 points) Let X, Y be jointly Gaussian random variables with mean zero and covariance matrix
Cov
( X Y
)
( 4 6 6 18
) .
You may express your answers in terms of the Φ function defined by Φ(u) =
∫ (^) u −∞ √^1 2 π e
−s^2 / (^2) ds.
(a) Find P [|X − 1 | ≥ 2].
(b) What is the conditional density of X given that Y = 3? You can either write out the density in full, or describe it as a well known density with specified parameter values.
(c) Find P [|X − E[X|Y ]| ≥ 1].
Problem 3 (8 points) Let (Xn, n ≥ 1) be a sequence of random variables and let X be a random variable such that P [X = c] = 1 for some constant c. Prove that if limn→∞ Xn = X d., then limn→∞ Xn = X p. That is, prove that convergence in distribution to a constant implies convergence in probability to the same constant.
Problem 4(8 points) Suppose in a given application a Kalman filter has been implemented to recur- sively produce ˆxk+1|k for k ≥ 0, as in class. Thus by time k, ˆxk+1|k, Σk+1|k, ˆxk|k− 1 , and Σk|k− 1 are already computed. Suppose that it is desired to also compute ˆxk|k at time k. Give additional equations that can be used to compute ˆxk|k. You may consult the attached sheet giving the definitions and equations for Kalman filtering. Be as explicit as you can, expressing any matrices you use in terms of the matrices of model or Kalman filter already considered.
Since ̂xk|k− 1 can be expressed as a linear transformation of (1, yk−^1 , yk), or equivalently as a linear transformation of (1, yk−^1 , y˜k),
x̂k+1|k = Fk ̂xk|k− 1 + Ê
[ xk+1 − Fk x̂k|k− 1 | yk−^1 , y˜k
]
. (6)
Since E[˜yk] = 0 and E[yk−^1 y˜kT ] = 0,
Ê [xk+1 − Fk x̂k|k− 1 | yk−^1 , ˜yk] = Ê [xk+1 − Fk x̂k|k− 1 | yk−^1 ] ︸ ︷︷ ︸
- Ê [xk+1 − Fk ̂xk|k− 1 | y˜k] 0
where the first term on the right side of (7) is zero by (4). Since xk+1 − Fk ̂xk|k− 1 and ˜yk are both mean zero,
Ê [xk+1 − Fkxk|k− 1 | y˜k] = Kk ˜yk (8)
where
Kk = Cov(xk+1 − Fk x̂ k|k− 1 , y˜k)Cov(˜yk)−^1. (9)
Combining (6), (7), and (8) yields the main Kalman filter equation
x̂k+1|k = Fk ̂xk|k+1 + Kk ˜yk. (10)
Taking into account the new observation ˜yk, which is orthogonal to the previous observations, yields a reduction in the covariance of error:
Σk+1|k = Σk+1|k− 1 − Cov(Kk y˜k). (11)
The Kalman filter equations (1), (5), and (3) follow easily from (10), (9), and (11), respectively. Some of the details follow. To convert (9) into (2), use
Cov(xk+1 − Fk x̂k|k− 1 , → y˜k) = Cov(Fk(xk − x̂ k|k− 1 ) + wk, HkT (xk − x̂ k|k− 1 ) + vk) (12) = Cov(Fk(xk − x̂ k|k− 1 ), HkT (xk − x̂k|k− 1 )) = FkΣk|k− 1 Hk
and
Cov(˜yk) = Cov(HkT (xk − x̂k|k− 1 ) + vk) = Cov(HkT (xk − x̂k|k− 1 )) + Cov(vk) = HkT Σk|k− 1 Hk + Rk
To convert (11) into (3) use (5) and
Cov(Kk y˜k) = KkCov(˜yk)KkT = Cov(xk+1 − Fk x̂ k|k− 1 )Cov(˜yk)−^1 Cov(xk+1 − Fk ̂xk|k− 1 )
This completes the derivation of the Kalman filtering equations.
