STEINs UNBIASED RISK ESTIMATOR (SURE) FOR OPTICAL FLOW Dr

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STEINs UNBIASED RISK ESTIMATOR (SURE) FOR OPTICAL FLOW Dr Powered By Docstoc
					 STEIN’s UNBIASED RISK
   ESTIMATOR (SURE)
   FOR OPTICAL FLOW

          Dr. Mingren Shi
    Mathematics and Computing
  University of Southern Queensland

OUTLINE

 What is optical flow?

 Overview: optical flow estimation

 Lucas-Kanade Method

 Stein’s Unbiased Risk Estimator (SURE)

 Conclusion



                   1
  What is optical flow?
Optical flow is the apparent motion of the bright-
ness/intensity patterns observed when a camera is
moving relative to the objects being imaged.
Optical flow equation (OFE)
• Let I = I(x1 , x2 , t) be image intensity (irradiance)
function at time t at the image point (x1 , x2 ).
• Let u = (u1 , u2 )T ≡ ( dx1 , dx2 )T be optical flow.
                           dt    dt
• Assume that the change of intensity of a particular
point in a moving pattern with the time is very small,
             dI     ∂I dx1     ∂I dx2    ∂I
   i.e. 0 ≈      =         +          +      .    (0.1)
             dt    ∂x1 dt     ∂x2 dt      ∂t
                ∂I       ∂I       ∂I
            or      u1 +     u2 +    ≈ 0.        (0.1)
                ∂x1      ∂x2      ∂t
• But the OFE (0.1) is ill-posed. Why?
    There are two variables (u1 , u2 ), but only one
    equation =⇒ it has infinity many solutions.

                u2                   u



                                      u1

             Fig. 1: Optical flow vector

                           2
• A constraint should be imposed on u.


  How to estimate optical flow?
• Many smoothing methods to estimate the optical
flow.
Lucas-Kanade method
       Assumption: the optical flow in a neigh-
    bourhood of the central pixel P is a constant.




                         3


                 3                    3

                             P



                         3




          Fig. 2: 3 × 3 neighbourhood size

                         3
  The smoothing parameter is
   the neighborhood size n (for (2n + 1) × (2n + 1)
pixels).
    Selection of this tuning parameter is difficult
   in general.
   But has a crucial rule and a profound effect on
   the results.
  • n too small =⇒ the numerical errors may be
    excessively amplified.
  • n too large =⇒ the solution may be not close
    to satisfying the real solution.
  • This is a tradeoff situation.
  • There is an optimal value of the n size.
  • Example: Divergent tree (The optical flow in
    each pixel is plotted in vector, see in Fig. 1.)
       image file: treed.mat
       data files: Treed1.dat - 11 × 11 (optimal)
                  Treed2.dat - 3 × 3 (too small)
                  Treed3.dat - 31 × 31 (too large)
                  Treed4.dat - SURE curve


                        4
  Lucas-Kanade Method
  Simplified natation

            ∂I              ∂I            ∂I
       It =     , Ix 1 =        , Ix2 =       ,
            ∂t             ∂x1            ∂x2     (1.0)
     P = (x1 , x2 ) : (central) position or pixel

Rewrite OFE (0.1) (wrt P ) as

      −It,P = u1,P Ix1 ,P + u2,P Ix2 ,P +           P
                                                        .   (1.1)

Moreover, let

      gP = (Ix1 ,P Ix2 ,P )T , uP = (u1,P u2,P )T ,
                      yP = −It,P ,
                                                 (1.2)
and (1.1) becomes

       T                                            T
 yP = gP uP +           P
                            = sP +   P
                                         ,   (sP = gP uP ). (1.3)

  The local constant LK estimator
• Assumptions
     NP is a pixel neighborhood of P .
     There are m = (2n + 1) × (2n + 1) pixels in NP .
A1    1, . . . ,   m
                       are independent.

                                 5
A2    = ( 1, . . . ,   N
                           )T ∼ N (0, σ 2 I)
• Definition of the estimator

       uP = arg min                        T
                                    (yQ − gQ uP )2 .         (1.4)
                             Q∈NP

• Euler equations wrt the estimator
                                      T
       uP   =0⇒                (yQ − gQ uP )gQ = 0,
                       Q∈NP
                         T                                 (1.5A)
     or (          gQ gQ )uP =                 yQ gQ
            Q∈NP                       Q∈NP

Therefore, we have
           −1                                               T
     uP = MP                 yQ gQ ,    MP =            gQ gQ .
                  Q∈NP                           Q∈NP
                                                           (1.5B)
• Remark NP = {P1 , P2 , . . . , Pm } ⊃ {P }
  gQ = (Ix1 ,Q , Ix2 ,Q )T and set
                Ix1 ,P1 . . . Ix1 ,Pm
   AT
    2×m =                                 = [gP1 . . . gPm ],
                Ix2 ,P1 . . . Ix2 ,Pm
     y = −(It,P1 . . . It,Pm )T = (yP1 . . . yPm )T .
  MP = AT A is a 2 × 2 matrix,                     yQ gQ = AT y.
                                          Q∈NP
                                            T
  (1.5A) will become AT AuP = A y which is the
“Regularized equation” of AuP = y.

                                  6
Stein’s Unbiased Risk Estimator (SURE)
• The SURE was originally used to estimate the
  mean of a multivariate normal distribution.
• We use it to estimate the tuning parameters in
  the context of image restoration.
Definition of SURE
• Minimising the expected value of the discrep-
  ancy (E is the expectation):

      R   = E(       |gP (uP − uP )|2 )
                       T
                 P

          = E(       |sP − sP |2 ),          T
                                      (sP = gP uP ),
                 P

          = E(       |(yP − sP ) − (yP − sP )|2 )
                 P
                                                        (2.1)
  The sum is for every pixel P in the image.
• Residual e and the error
       e = (yP1 − sP1 , . . . , yPN − sPN )T ,
                                                        (2.2)
         = (yP1 − sP1 , . . . , yPN − sPN )T .

• The discrepancy measure (2.1) is

  R = E||e − ||2 = E||e||2 − 2E(          T
                                              e) + E|| ||2 .
                                                        (2.3)

                          7
   An estimator of R
(1) An unbiased estimator of E||e||2 is ||e||2 , which
      can be calculated directly once the flow estimate
      u is obtained.
                                                   2            2
(2) Since E(              P
                              ) = 0, E(            P
                                                     ) = E(     P
                                                                  ) − [E(   P
                                                                                )]2 ≡
      σ 2 , we have
      E( 2 ) =            E(       2
                                   P
                                     ) = N · E(         2
                                                        P
                                                          ) = N σ2 .
                      P
(3)    As usual, we must work on the middle term
           T
      E(       e).
   Calculation of the middle term
Stein’s Lemma                      Given the Gaussian assumption
(A1 & A2) and u is weakly differentiable with re-
spect to the data y, we have
                              T            2     ∂e
                     E(           e) = σ E[trace( )].                           (2.4)
                                                 ∂y
Proof      The distribution density is
      p(y|u) ≡ p( ) ≡ p( 1 , . . . ,                   N
                                                            )
                          N
                     =            p( i )   by independent
                          i=1
                           N                                     2
                                                                                (2.5)
                     =            (2πσ 2 )−1/2 exp(−             i
                                                                2σ 2   )
                          i=1
                                            1
                     = (2πσ 2 )−N/2 exp[−( 2σ2 )|| ||2 ],

                                               8
Furthermore, since             || ||2 = 2 ,

        ∂p
        ∂y    = p(y|u)[− 2σ2 (2 )] = − p(y|u) ,
                          1
                                         σ2
                                                                (2.6)
               ⇒     ∂p
                   ( ∂y )T e   =    − p(y|u) ( T e).
                                        σ2

Thus,
        T
  E(        e) = [p(y|u)( T e)]dy
                         ∂p
               = −σ 2 [( ∂y )T e] dy
                                        N
                     2                        ∂ei
                =σ        [p(y|u)             ∂yi ]dy
                                        i=1
                (integration by parts & p(±∞) = 0)
                                      ∂e
                = σ 2 [p(y|u)trace( ∂y )] dy
                               ∂e
                = σ 2 E[trace( ∂y )].
                (by the definition of E)
                                                (2.7)
The calculation of ∂e/∂y


    ∂eP            ∂                              ∂ sP
    ∂yP       =   ∂yP   (yP − sP ) = 1 −          ∂yP
                          ∂      T             T         ∂ uP
              =1−        ∂yP   (gP uP ) = 1 − gP         ∂yP
                                                                (2.8)
                     T ∂    −1
              = 1 − gP ∂y [MP                        yQ gQ ]
                                P
                                              Q∈NP
                     T
              = 1 − gP M −1 gP . from (1.5B)

                                    9
  From the above, an unbiased estimate of the risk
is given by:
                             ∂e
 R    ≈ ||e||2 − 2σ 2 trace( ∂y ) + N σ 2
                                       T
      = [(yP − sP )2 − 2σ 2 (1 − gP M −1 gP ) + σ 2 ]
          P
                        (by (2.2), (2.8))
      =          2
              [(yP − 2yP sP + s2 )
                               P
          P
             T
      +2σ 2 gP M −1 gP − σ 2 ]
                                                 (2.9)
             2
We can drop yP and −σ 2 since both terms do not
depend on the neighborhood size. So we can replace
R by R as follows:

                      R=        RP ,
                            P                   (2.11)
                               T  −1
     RP = −2yP sP + s2 + 2σ 2 gP MP gP .
                     P




                             10
    CONCLUSION

    The Optical flow equation is ill-posed. An extra
constraints should impose on the optical flow vector
u


    The smoothing parameter for the local constant
Lucas-Kanade method is the neighbourhood size n
which is tradeoff.


    Stein’s Unbiased Risk Estimator can be used to
find the optimal value of the tuning parameter n.




                         11
      Questions?

     Comments?

Constructive Criticism?

     Suggestions?


       Thank you




          12

				
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Description: STEINs UNBIASED RISK ESTIMATOR (SURE) FOR OPTICAL FLOW Dr