ISI Platinum Jubilee, Jan 1-4, 2008
Roy’s union-intersection
principle, random fields,
and brain imaging
Keith Worsley,
McGill
Jonathan Taylor,
Stanford and Université de Montréal
Sankhya (1937) 3:65-72
Deformation Based Morphometry (DBM)
(Tomaiuolo et al., 2004)



n1 = 19 non-missile brain trauma patients, 3-14 days in coma,
n2 = 17 age and gender matched controls
Data: non-linear vector deformations needed to warp each MRI
to an atlas standard


Y1i(s) = deformation vector of trauma patient i at point s
Y2i(s) = deformation vector of control patient i at point s

Locate damage: find regions where
deformations are different

Test statistic: Hotelling’s T2 (1931)


T(s)2 = (Y1.(s) – Y2.(s))’Σ-1 (Y1.(s) – Y2.(s))
P.C. Mahalanobis with Harold Hotelling,
at Gupta Nivas, 1939
Multivariate linear models
for random field data
Y(s)n£q is the observations £ variables data matrix at point s 2 S ½ <D .
At every point s we have a multivariate linear model:
Y(s)n£q = Xn£p B(s)n£q + E(s)n£q ;
E(s)n£q » N(0; I
§):
We detect sparse s where B(s) 6= 0 by testing H0 : B(s) = 0 at every point s.
Test statistics T (s):
# regressors
p=1
p>1
q=1
T
F
# variables
q>1 Hotelling’s T2
Need null distribution: P(maxs2S T (s) ¸ t)
Wilks’ Λ
Pillai’s trace
Roy’s max root
etc.
EC heuristic
At high thresholds t, the excursion set Xt = fs : T (s) ¸ tg is either one
connected component (EC = 1) or empty (EC = 0), so
µ
¶
P max T (s) ¸ t ¼ E(EC(S \ Xt )):
s2S
Theorem (1981, 1995)
If T (s) is a smooth isotropic random ¯eld then
E(EC(S \ Xt )) =
X
D
d=0
¹d (S)½d (t):
E(EC(S \ Xt )) =
X
D
¹d (S)½d (t)
d=0
Intrinsic volume ¹d (S)
EC density ½d (t)
For S with smooth boundary @S
that has curvature matrix C(s),
Morse theory:
X
\
EC(S Xt ) =
1fT (s)¸tg 1f@T (s)=@s=0g
¡((D ¡ d)=2)
¹d (S) =
¡
Z 2¼(D d)=2
£
detrD¡1¡d fC(s)gds;
@S
and ¹D (S) = jS j. For D = 3,
¹0 (S) = EC(S)
¹1 (S) = 2 Diameter(S)
¹2 (S) = 1 Surface area(S)
2
¹3 (S) = Volume(S):
s
µ
¶
£ sign ¡ @ 2 T
+ boundary
@s@s0
µ
µ
¶
2
@ T
½D (t) = E 1fT ¸tg det ¡
@s@s0
¯
¶ µ
¶
¯ @T
@T
¯
P
=
0
=0
¯ @s
@s
For a Gaussian random ¯eld, ¸ = Sd( @Z ),
@s
µ ¡
¶
¸ @ d
p
P(Z ¸ t)
½d (t) =
2¼ @t
After lots of messy algebra …
EC densities are known for Â2 , T , F ¯elds (1994), and Hotelling0 s T 2 , º df
(1999):
½0 (t) =
½1 (t) =
½2 (t) =
½3 (t) =
Z
1
¡( º+1 )
¡ º+1 q¡2
2
(1
+
u)
u 2 du;
2
¡q+1
q
º
¡( )¡(
)
t
2
2
¼ ¡ 12 ¡( º+1 )
¡ º ¡1 q¡1
2
(1
+
t)
t 2 ;
2
¡q+2
q
º
¡( )¡(
)
2
2
µ
¶
¡
¡
º+1
1
¼ ¡(
)
q 1
¡ º ¡1 q¡2
¡
2
(1 + t) 2 t 2
t
;
º ¡q+1
¡( q )¡( º ¡q+1 )
2
2
µ
¶
¡
¡
¡
¡
3
º+1
¼ 2 ¡(
)
2q 1
(q 1)(q 2)
¡ º ¡1 q ¡3
¡
2
2
(1 + t) 2 t 2
t
t+ ¡
:
º ¡q
(º q + 2)(º ¡ q)
¡( q )¡( º ¡q )
2
2
Last case
# regressors
p=1
p>1
q=1
# variables
T✔
F✔
q>1 Hotelling’s T2✔ Wilks’ Λ
Pillai’s trace
Roy’s max root
etc.
?
We shall now ¯nd a P-value approximation (but not quite the EC density) for
the maximum Roy0 s maximum root,
T (s) =maximum eigenvalue of
Y(s)0 X(X0 X)¡1 X0 Y(s)(Y(s)0 (I ¡ X(X0 X)¡1 X0 )Y(s))¡1 :
The above messy algebra is just too complicated. Instead . . .
Roy’s union-intersection principle
Make it into a univariate linear model by multiplying by vector uq£1 :
(Y(s)u)n£1 = Xn£p (B(s)u)q£1 + (E(s)u)n£1 ;
H0u : (B(s)u)q£1 = 0
Let F (s; u) be the usual F-statistic for testing H0u . Let °q ½ <q be the unit
q-sphere. Roy0 s maximum root is
T (s) = maxu2° F (s; u):
q
Now F (s; u) is an F-¯eld in search region S °q and we already know the EC
density of the F-¯eld, ½d (F ¸ t) so
µ
¶
µ
¶
P max T (s) ¸ t = P
max F (s; u) ¸ t
s2S
s2S;u2°q
¼
1
2
D+q
X
¹d (S
° )½ (F ¸ t)
q d✔
d=0
Why ½? F(s,u)=F(s,-u)
=
1
2
X
D
d=0
¹d (S)
q
X
k=0
Almost the EC
density of the
Roy’s maximum
root field
¸ t)
¹k (°q )½d+k
(F
✔
Non-negative least squares
random field theory
fMRI data: 120 scans, 3 scans hot, rest, warm, rest, …
First scan of fMRI data
1000
Highly significant effect, T=6.59
500
hot
rest
warm
890
880
870
0
0
100
200
300
No significant effect, T=-0.74
820
hot
rest
warm
T statistic for hot - warm effect
5
800
0
100
0
100
0
-5
T = (hot – warm effect) / S.d.
~ t110 if no effect
200
Drift
300
810
800
790
200
Time, seconds
300
Linear model regressors
Alternating hot and warm stimuli separated by rest (9 seconds each).
2
1
0
-1
0
50
100
150
200
250
Hemodynamic response function (HRF)
300
350
0.4
Delays and disperses
the stimuli by ~6s
0.2
0
-0.2
0
50
Regressors x(t) = stimuli * HRF, sampled every 3 seconds
2
1
0
-1
0
50
100
150
200
Time, seconds
250
300
350
Linear model for fMRI time series
with AR(p) errors
Y (t) = x(t)¯?+ z(t)°? + ²(t)
²(t) = a1 ²(t ¡ 1) + ¢ ¢ ¢ + ap ²(t ¡ p) + ¾W N (t)
?
?
?
t = time
Y (t) = fMRI data
x(t) = stimuli convolved with HRF
z(t) = drift etc:
²(t) = error
W N (t) » N(0; 1) independently
unknown
parameters
Allowing for unknown delay of the HRF
Replace x(t) by two responses with extreme HRF delays
x1 (t) = x(t) with HRF shifted by + 2 seconds;
x2 (t) = x(t) with HRF shifted by ¡ 2 seconds;
Y (t) = x1 (t)¯1 + x2 (t)¯2 + z(t)° + ²(t):
When the coe±cients are non-negative
¯1 ¸ 0;
¯2 ¸ 0;
we get a range of responses in between:
x(t)
2
1
x2(t)
x1(t)
stimulus
0
-1
10
20
30
40
50
Time, t (seconds)
60
70
Non-negative coefficients is a cone alternative
Recall model is
Y (t) = x1 (t)¯1 + x2 (t)¯2 + z(t)° + ²(t)
¯ ¸ 0; ¯ ¸ 0:
1
2
Let z1 , z2 be an orthogonal basis for x1 , x2 .
Then E(Y ) falls inside a cone alternative:
Cone angle θ =
angle between
x1 and x2
x1(t)
z2
Cone
alternative
Null 0
β1 ≥ 0
β2 ≥ 0
z1
x2(t)
Example of three extremes
Y (t) = x1 (t)¯1 + x2 (t)¯2 + x3 (t)¯3 + z(t)° + ²(t);
¯ ¸ 0; ¯ ¸ 0; ¯ ¸ 0;
1
2
3
z1 ; z2 ; z3 orthogonal basis for x1 ; x2 ; x3 .
4
x3(t)
spread
4 seconds
3D cone
0
25
4
x1(t)
standard
z3
z1
z2
x2(t)
delayed
4 seconds
General non-negative least squares problem
In general we may have p constrained regressors and q unconstrained regressors.
The constrained regressors could be the extremes of the HRF, e.g. min/max
latency shift, min/max spread, etc. The model is then all HRFs in between. In
vector form:
Yn£1 = Xn£p ¯p£1 + Zn£q °q£1 + ²n£1
¯p£1 ¸ 0 (component ¡ wise)
°q£1 unconstrained
²n£1 » N(0n£1 ; In£n ¾ 2 ) (without loss of generality)
We might also have far more regressors than actual observations(!), i.e. p >> n.
Footnote: Woolrich et al. (2004) replace “hard” constraints by “soft” constraints
through a prior distribution on β, taking a Bayesian approach.
Pick a range of say p = 150 plausible values of the non-linear parameter
ν: ν1,…,νp
Fitting the NNLS model
Simple:
• Do “all subsets” regression
• Throw out any model that does not satisfy the non-negativity constraints
• Among those left, pick the model with smallest error sum of squares
For larger models there are more efficient methods e.g. Lawson & Hanson
(1974).
The non-negativity constraints tend to enforce sparsity, even if regressors are
highly correlated (e.g. PET).
Why? Highly correlated regressors have huge positive and negative
unconstrained coefficients – non-negativity suppresses the negative ones.
Example:
n=20, p=150, but surprisingly it does not overfit
30
Y
Yhat
Yhat component 1
Yhat component 2
25
Tracer
20
b̄ = 107:4
b̄40 = 46:9
41
15
10
5
0
0
1000
2000
b̄ = 5:4
b̄86 = 71:7
87
3000
Time
4000
rest:
b̄ = 0
j
5000
6000
Tend to get sparse pairs of adjacent regressors, suggesting best regressor
is somewhere inbetween.
P-values for testing the cone alternative?
!
H0 : ¯p£1 = 0p£1
error sum of squares = SSE0
H1 : ¯p£1 ¸ 0p£1 (component ¡ wise) ! error sum of squares = SSE1
Likelihood ratio test statistic is the Beta-bar statistic, equal to the coe±cient of
determination or multiple correlation R2 :
¡ SSE
SSE
0
1
¹=
B
SSE0
Null distribution if there are no constraints, i.e. H1 : ¯p£1 6= 0p£1 :
¡
¡
¢¸ ¢
¡
¸
¹
P(B t) = P Beta p ; º p
t ; º =n¡q
2
2
Null distribution with constraints is a weighted average of Beta distributions,
hence the name Beta-bar (Lin & Lindsay, Takemura & Kuriki, 1997):
¹ ¸ t) =
P(B
X
º
j=1
¡
¡
¢¸ ¢
¡
j
º
j
wj P Beta ;
t
2
=1 when j=ν
2
wj = P(#funconstrained b̄0 s > 0g = j)
P-values¹ for PET data at a single voxel
Observed B
is t = 0:9524, º = 20.
p = 2: Cone weights:
w1 = ½, w2 = θ/2π
¹ weights w
Easiest way to ¯nd the B
j
is by simulation (10000 iterations):
-8
0.5
2
0.4
wj
1.5
x 10
¡
¡
¢¸ ¢
¡
j
º
j
P Beta ;
t
2
0.3
1
2
0.2
0.5
0.1
0
0
1
2
3
4
5
0
6
j
¹ ¸ t) =
P(B
X
º
j=1
0
1
2
3
4
j
¡
¡
¢¸ ¢
¡
wj P Beta j ; º j
t = 2:35 £ 10¡12
2
2
5
6
The Beta-bar random field
Recall that at a single point (voxel):
¹ ¸ t) =
P(B
X
º
j=1
¡
¡
¢¸ ¢
¡
j
º
j
wj P Beta ;
t
2
2
Recall that if T (s); s 2 S ½ <D , is an isotropic random ¯eld:
µ
¶
X
D
¸
¼
\
f
¸
g
P max T (s) t
E(EC(S
s : T (s) t )) =
¹d (S)½d (T ¸ t)
s2S
d=0
with ½0 ´ P. Taylor & Worsley (2007):
¹ ¸ t) =
½d (B
X
º
j=1
¡
¡
¢
¢
¡j ¸
j
º
wj ½d Beta ;
t
2
=1 when j=ν
2
From well-known
EC
µ From simulations
¶ X
µ
¶
Same
linear
combination!
¡
¢ densities
at a single¸
voxel ¼ º
of F field
¡
¸
¹
j
º
j
P max B(s) t
wj P max Beta ;
(s) t
s2S
j=1
s2S
2
2
Roy’s union-intersection principle
Yn£1 = Xn£p ¯p£1 + ²n£1 ; ¯p£1 ¸ 0; (component ¡ wise)
» N(0
²
;I
¾ 2 ): (without loss of generality)
n£1
n£1
n£n
Write the model in terms of the orthonormalised regressors:
Yn£1 = (Zu)n£1 ® + ²n£1 ;
0
® ¸ 0;
X)¡1=2 ;
Zn£p = X(X
½
¾
0
1=2
(X X) ¯
¸ 0 ½ ° ½ <n :
up£1 2 U = jj 0
:
¯
p
(X X)1=2 ¯ jj
c ®
b(u)=Sd(
b(u)) be the usual T-statistic for
Let T (u) = ®
testing H0u : ® = 0. Roy0 s UIP gives the test statistic:
T¹ = max T (u);
u2U
¹=
B
T¹2
:
¡
¹
n 1 + T2
x1(t)
z2
Cone
u
0
1
z1
U
x2(t)
Roy’s union-intersection principle
Can we use the same trick of adding U to the search space?
Let0 s put back the parameter s. The T-bar statistic is
T¹(s) = max T (s; u); T (s; u) »H tn¡1 ;
0
u2U
µ
¶
µ
¶
P max T¹(s) ¸ t = P
max T (s; u) ¸ t
s2S
s2S;u2U
¼
D+q
X
¹d (S
U )½✔d (T ¸ t)
d=0
=
X
D
d=0
¹d (S)
q
X
¸ t):
¹k (U )½d+k
✔ (T
k=0
Sadly this doesn0 t work: although T (s; u) is a T-¯eld in s for each u, it is not
b(s; u) is a Gaussian random ¯eld in
a T-¯eld in (s; u), despite the fact that ®
c ®
b(s; u))2 is not a Â2 ¡ ¯eld in (s; u).
(s; u). The reason is that (n ¡ 1)Sd(
n 1
Â
¹=
max Z1 cos µ + Z2 sin µ
0·µ·¼=2
Functions of Gaussian fields:
Z1~N(0,1)
Example:
3
Z2~N(0,1)
s2
2
1
0
-1
-2
-3
Excursion sets,
Xt = fs : Â
¹ ¸ tg
Search
Region,
S
Rt = fZ : Â
¹ ¸ tg
s1
Rejection regions,
Z2
Cone
2
alternative
4
0
2
Z1
Null
3
1
-2
-2
0
2
Threshold t
Euler characteristic heuristic again
Search Region, S
Excursion sets, Xt
EC= #blobs - # holes
= 1
7
6
5
2
1
1
Euler characteristic, EC
10
Heuristic :
¸ t)
P(max Â(s)
¹
Observed
8
0
s2S
6
¼ E(EC) = 0:05
Expected
4
) t = 3:75
2
0
-2
0
0.5
EXACT!
1
1.5
2
E(EC(S \ Xt )) =
X
D
d=0
2.5
3
L (S)½ (R )
d
d
t
3.5
4
Threshold, t
E(EC(S \ Xt )) =
Beautiful symmetry:
X
D
µ
L (S)½ (R )
d
d
t
@Z
¸ = Sd
@s
d=0
Lipschitz-Killing curvature Ld (S)
Steiner-Weyl Tube Formula (1930)
¶
EC density ½d (Rt )
Taylor Gaussian Tube Formula (2003)
• Put a tube of radius r about the search region λS and rejection region Rt:
Z2~N(0,1)
14
r
12
10
Rt
Tube(λS,r)
8
Tube(Rt,r)
r
λS
6
t-r
t
Z1~N(0,1)
4
2
2
4
6
8 10 12 14
• Find volume or probability, expand as a power series in r, pull off1coefficients:
jTube(¸S; r)j =
X
D
d=0
¼d
L
P(Tube(Rt ; r)) =
¡d (S)r d
D
¡(d=2 + 1)
X (2¼)d=2
d!
d=0
½d (Rt )rd
Lipschitz-Killing
curvature Ld (S)
of a triangle
r
Tube(λS,r)
λS
¸ = Sd
µ
@Z
@s
¶
Steiner-Weyl Volume of Tubes Formula (1930)
Area(Tube(¸S; r)) =
X
D
¼ d=2
L
¡d (S)r d
D
¡(d=2 + 1)
d=0
= L2 (S) + 2L1 (S)r + ¼ L0 (S)r2
= Area(¸S) + Perimeter(¸S)r + EC(¸S)¼r2
L (S) = EC(¸S)
0
L (S) = 1 Perimeter(¸S)
1
2
L (S) = Area(¸S)
2
Lipschitz-Killing curvatures are just “intrinisic volumes” or “Minkowski functionals”
in the (Riemannian) metric of the variance of the derivative of the process
Lipschitz-Killing curvature Ld (S) of any set
S
S
S
¸ = Sd
Edge length × λ
12
10
8
6
4
2
.
.. . .
.
. . .
.. . .
.. . .
.
. . . .
. . . .
. . . .
.. . .
. . .
... .
.
4
..
.
.
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6
..
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8
..
. .
. ...
. ..
. . .
. .
. . .....
. . .
. ....
..
..
10
µ
@Z
@s
¶
of triangles
L (Lipschitz-Killing
²) = 1, L (¡) curvature
L (N
=
1,
)=1
0
0
0
L (¡) = edge length, L (N) = 1 perimeter
1
2
L1 (N) = area
2
P Lcurvature
P L
Lipschitz-Killing
union
L
² ¡ Pof L
¡ of triangles
N
(S) = P² 0 ( )
¡ 0( ) +
P
L (S) =
L (¡) ¡
L (N)
¡
N 1
L1 (S) = P L 1(N)
2
N 2
0
N
0
( )
s2
Z~N(0,1)
Non-isotropic data?
µ
¸ = Sd
3
@Z
@s
¶
2
1
0.14
0.12
0
-1
-2
s1
0.1
0.08
0.06
-3
. . . .. .
12we warp
..
• Can
to isotropy? i.e. multiply edge lengths by λ?
... . .the
. . data
•
•
•
.
. . . . . . . .
10 .
. . . . . . . ...
. . no,
Globally
. . but
. . locally
. . . . yes, but we may need extra dimensions.
.
8 . . . . . . . . . .
. .
. . . . . . . Theorem:
Nash
Embedding
#dimensions ≤ D + D(D+1)/2; D=2: #dimensions
6 . . . . . . . . . .....
.. . . . . . . . .
. . . . Euclidean
....
4 idea:
. . . replace
Better
distance by the variogram:
... . . . . ...
. . . . .
d(s1, s2)2 = Var(Z(s1) - Z(s2)).
2
4
6
8
10
≤ 5.
Non-isotropic data? Use Riemannian metric
µ
¶ of Var(∇Z)
¸ = Sd
Z~N(0,1)
s2
3
@Z
@s
2
1
0.14
0.12
0
-1
-2
Edge length × λ
12
10
8
6
4
2
..
.
.
.
. .
. .
.
.. .
.
.. .
.
.
.
.
.
.
.
. . .
.
. . .
.. .
.
.
. .
...
.
. .
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4
6
..
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8
.
..
. .
. ...
. ..
. . .
. .
. . .....
. . .
. ....
...
10
s1
0.1
0.08
0.06
-3
of triangles
L (Lipschitz-Killing
²) = 1, L (¡) curvature
L (N
=
1,
)=1
0
0
0
L (¡) = edge length, L (N) = 1 perimeter
1
2
L1 (N) = area
2
P Lcurvature
P L
Lipschitz-Killing
union
L
² ¡ Pof L
¡ of triangles
N
(S) = P² 0 ( )
¡ 0( ) +
P
L (S) =
L (¡) ¡
L (N)
¡
N 1
L1 (S) = P L 1(N)
2
N 2
0
N
0
( )
Estimating Lipschitz-Killing curvature Ld (S)
We need independent & identically distributed random fields
e.g. residuals from a linear model
Z1
Z2
Z3
Z4
Replace coordinates of
the triangles in S ½ <2
by normalised residuals
Z
jjZ jj ;
Z = (Z1 ; : : : ; Zn ) 2 <n :
Taylor & Worsley, JASA (2007)
Z5
Z6
Z7
Z8
Z9 … Zn
of triangles
L (Lipschitz-Killing
²) = 1, L (¡) curvature
L (N
=
1,
)=1
0
0
0
L (¡) = edge length, L (N) = 1 perimeter
1
2
L1 (N) = area
2
P Lcurvature
P L
Lipschitz-Killing
union
L
² ¡ Pof L
¡ of triangles
N
(S) = P² 0 ( )
¡ 0( ) +
P
L (S) =
L (¡) ¡
L (N)
¡
N 1
L1 (S) = P L 1(N)
2
N 2
0
N
0
( )
E(EC(S \ Xt )) =
Beautiful symmetry:
X
D
µ
L (S)½ (R )
d
d
t
@Z
¸ = Sd
@s
d=0
Lipschitz-Killing curvature Ld (S)
Steiner-Weyl Tube Formula (1930)
¶
EC density ½d (Rt )
Taylor Gaussian Tube Formula (2003)
• Put a tube of radius r about the search region λS and rejection region Rt:
Z2~N(0,1)
14
r
12
10
Rt
Tube(λS,r)
8
Tube(Rt,r)
r
λS
6
t-r
t
Z1~N(0,1)
4
2
2
4
6
8 10 12 14
• Find volume or probability, expand as a power series in r, pull off1coefficients:
jTube(¸S; r)j =
X
D
d=0
¼d
L
P(Tube(Rt ; r)) =
¡d (S)r d
D
¡(d=2 + 1)
X (2¼)d=2
d!
d=0
½d (Rt )rd
EC density ½d (Â
¹ ¸ t)
of the Â
¹ statistic
Z2~N(0,1)
Tube(Rt,r)
r
t-r
Rejection region
Rt
t
Z1~N(0,1)
Taylor’s Gaussian Tube Formula
(2003)
1
P (Z1 ; Z2 2 Tube(Rt ; r) =
X (2¼)d=2
d!
½d (Â
¹ ¸ t)rd
d=0
(2¼)1=2 ½1 (Â
¹
¸ t)r + (2¼)½ (Â
¸ t)r2 =2 + ¢ ¢ ¢
= ½0 (Â
¹ ¸ t) +
¹
2
Z 1
=
(2¼)¡1=2 e¡z2 =2 dz + e¡(t¡r)2 =2 =4
t¡r
½0 (Â
¹ ¸ t) =
Z
t
1
(2¼)¡1=2 e¡z2 =2 dz + e¡t2 =2 =4
½1 (Â
¹ ¸ t) = (2¼)¡1 e¡t2 =2 + (2¼)¡1=2 e¡t2 =2 t=4
½ (Â
¹ ¸ t) = (2¼)¡3=2 e¡t2 =2 t + (2¼)¡1 e¡t2 =2 (t2 ¡ 1)=8
2
..
.
¹ ¸ t) of the B
¹ statistic
EC density ½d (B
Recall that at a single point (voxel):
¹ ¸ t) =
P(B
X
n
j=1
¡
¡
¢¸ ¢
¡
j
º
j
wj P Beta ;
t
2
2
Recall that if F (s); s 2 S ½ <D , is an isotropic random ¯eld:
µ
¶
X
D
¸
¼
\
f
¸
g
L (S)½ (F ¸ t)
P max Z(s) t
E(EC(S
s : Z(s) t )) =
d
d
s2S
d=0
with ½0 ´ P. Taylor & Worsley (2007):
¹ ¸ t) =
½d (B
X
n
j=1
¡
¡
¢
¢
¡j ¸
j
º
wj ½d Beta ;
t
2
2
From well-known
EC
µ From simulations
¶ X
µ
¶
linear combination!
n
¡
¢ densities
at a single¸
voxel ¼ Same
of F field
¡
¸
¹
j
º
j
P max B(s) t
wj P max Beta ;
(s) t
s2S
j=1
s2S
2
2
Power?
S = 1000cc brain, FWHM = 10mm, P = 0.05
Event
Block (20 seconds)
1
1
Cone angle θ = 78.4o
Cone angle θ = 38.1o
0.9
0.9
T-test on β1
0.8
0.8
0.7
0.6
0.6
Power of test
F-test
on
0.5 (β , β )
1
2
0.4
Cone
weights:
w1 = ½
w2 = θ/2π
0.5
0.4
0.3
0.2
0
0
0.3
x1(t)
0
0.2
x2(t)
-0.5
0.1
2
0
2
Shift d of HRF (seconds)
3
0
0
-2
0.1
20
40
Time t (seconds)
1
Response
0.5
Response
Power of test
Beta-bar test
0.7
0
0
20
40
Time t (seconds)
1
2
Shift d of HRF (seconds)
3
Cross correlation random field
Let X(s), s 2 S ½ <M , and Y (t), t 2 T ½ <N be n £ 1 vectors of
Gaussian random ¯elds. De¯ne the cross correlation random ¯eld as
X(s)0 Y (t)
C(s; t) = max
u;v (X(s)0 X(s)
P
µ
(t))1=2
:
¶
max C(s; t) ¸ c ¼ E(EC fs 2 S; t 2 T : C(s; t) ¸ cg)
s2S;t2T
=
dim(S)
X dim(T
X)
i=0
½ij (C ¸ c) =
Y
(t)0 Y
2n¡2¡h (i
L (S)L (T )½ (C ¸ c)
i
j
ij
j=0
¡1)=2c
¡ 1)!j! b(hX
¼h=2+1
k=0
(¡1)k ch¡1¡2k (1 ¡ c2 )(n¡1¡h)=2+k
X
k X
k
l=0 m=0
¡( n¡i + l)¡( n¡j + m)
2
2
:
¡
¡
¡
¡
l!m!(k l m)!(n 1 h + l + m + k)!(i ¡ 1 ¡ k ¡ l + m)!(j ¡ k ¡ m + l)!
Maximum canonical cross correlation random field
Let X(s)n£p , s 2 S ½ <M , and Y(t)n£q , t 2 T ½ <N be matrices of Gaussian
random ¯elds. De¯ne the maximum canonical cross correlation random ¯eld as
u0 X(s)0 Y(t)v
C(s; t) = max
u;v (u0 X(s)0 X(s)u
v 0 Y(t)0 Y(t)v)1=2
;
the maximum of the canonical correlations between X and Y, de¯ned as the
singular values of (X0 X)¡1=2 X0 Y(Y 0 Y)¡1=2 .
P
µ
max C(s; t) ¸ c
s2S;t2T
¶
¼
1
2
X
M
L (S)
i
i=0
where
L (° ) =
k
p
X
N
L (T )
j
j=0
p
X
L (° )
k
p
k=0
¡
2k+1 ¼ k2 ¡ p+1
³
´2
¡
¡
k! p 1 k !
2
q
X
l=0
¢
if p ¡ 1 ¡ k is even, and zero otherwise, k = 0; : : : ; p ¡ 1.
L (° )½
¸ c)
(C
l
q i+k;j+l
Mann-Whitney random field
MW = sum of ranks of n/2 random fields, n=10
2.5
2
50
1.5
1
100
0.5
0
150
-0.5
-1
-1.5
200
-2
-2.5
250
50
100
150
200
250