Download Solutions to ECE544 Homework #3: Statistical Image and Video Processing - Problems 1 and 2 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
ECE
Statistical Image and Video Processing
Homework #3 Solutions
Solution to Problem 1
Consider the following HMM. The hidden sequence S 0 , S 1 , S 2 , S 3 is a homogeneous
Markov chain with alphabet Λ = { 0 , 1 } and transition probability matrix Q =
The observed variables are X = {X 0 , · · · , X 7 } ∈ Λ^8. The conditional pmf for X given S is given by
p(x|s) =
∏^3
i=
Q(x 2 i|si)Q(x 2 i+1|si).
Give a computationally efficient algorithm for evaluating p(x 7 |s 0 ).
Solution: write
p(x 7 |s 0 ) =
s 1
Q(s 1 |s 0 )
s 2
Q(s 2 |s 1 )
s 3
Q(s 3 |s 2 )Q(x 7 |s 3 )
x 6
Q(x 6 |s 3 ) · · ·
x 1
Q(x 1 |s 0 )
x 0
Q(x 0 |s 0 )
Observe that the sums over x 0 · · · x 6 are all equal to 1, hence
p(x 7 |s 0 ) =
s 1
Q(s 1 |s 0 )
s 2
Q(s 2 |s 1 )
s 3
Q(s 3 |s 2 )Q(x 7 |s 3 ) ︸ ︷︷ ︸ m(s 2 ,x 7 ) =
s 1
Q(s 1 |s 0 )
s 2
Q(s 2 |s 1 )m(s 2 , x 7 ) ︸ ︷︷ ︸ m(s 1 ,x 7 ) =
s 1
Q(s 1 |s 0 )m(s 1 , x 7 )
i.e., p(x 7 |s 0 ) is computed by successive evaluation of the above messages.
Solution to Problem 2
The global orientation of the hand is represented by a variable O, all fingers have prespec- ified lengths, and the angle subtended by fingers 1 and 5 (the pinkie and the thumb)
is denoted by A. The orientation of the five fingers is represented by angular vari- ables F 1 , · · · , F 5. In this simplified model, the hand is parameterized by the 7 variables F 1 , · · · , F 5 , O, A.
These variables are modeled as random with the following joint pdf:
p(f 1 , · · · , f 5 , o, a) =
Z
ψ 15 (f 1 , f 5 |o, a)ψ 12 (f 1 , f 2 )ψ 23 (f 2 , f 3 )ψ 34 (f 3 , f 4 )ψ 45 (f 4 , f 5 )p(o)p(a).
- A graphical representation of this model is given below.
The marginal p(f 3 ) may be evaluated by eliminating o, a, f 1 , f 2 , f 5 , f 4 in that order:
p(f 3 ) =
Z
f 4
ψ 34 (f 3 , f 4 )
f 5
ψ 45 (f 4 , f 5 )
f 2
ψ 23 (f 2 , f 3 )
f 1
ψ 12 (f 1 , f 2 )
×
o,a
ψ 15 (f 1 , f 5 |o, a)p(o)p(a).
- This model appears physiologically plausible because muscle configuration controls the local interactions between adjacent fingers, as you may verify experimentally. Note that the thumb is quite independent of the other fingers.
- Assume now that O = o∗^ and A = 10 (degrees) with probability 1, and that ψi,i+1(fi, fi+1) = exp{fi − fi+1} (^1) {fi<fi+1} : i = 1, 2 , 3
= exp{
(fi − fi+1)} (^1) {fi<fi+1} : i = 4
ψ 15 (f 1 , f 5 |o∗, 10) = e−^10
(^6) |f 1 +5|− 106 |f 5 − 5 | .
With high probability, F 1 and F 5 are very close to -5 and 5, respectively. We can analytically evaluate E[F 3 ] because the distributions are exponential. First we write
p(f 3 ) =
df 1 df 2 df 4 df 5 p(f 1 · · · f 5 |o∗, A = 10)
4 Z
∫ (^) f 2
−∞
df 1 ef^1 −f^2 e−^10 (^6) |f 1 +5|
∫ (^) f 3
−∞
df 2 ef^2 −f^3
f 3
df 4 ef^4 −f^3
f 4
df 5 e
1 4 (f^4 −f^5 )^ e−^106 |f^5 +5|
10 −^12
Z
∫ (^) f 2
−∞
df 1 ef^1 δ(f 1 + 5)
∫ (^) f 3
−∞
df 2
f 3
df 4 e−^
3 4 f^4
f 4
df 5 e−^
1 4 f^5 δ(f 5 − 5)
10 −^12 e−^5 −^ (^14 )
Z
∫ (^) f 3
− 5
df 2
f 3
df 4 e−^
(^34) f 4
4 3 10
(^6) e− (^154) Z
(f 3 + 5)(e−^
(^34) f 3 − e−^
(^154) )
(f 3 + 5)(e−^
(^34) f 3 − e−^
(^154) ) ∫ (^5) − 5 (f^3 + 5)(e
− 34 f (^3) − e− (^154) ) df 3
