Random shapes in brain
mapping and astrophysics
using an idea from geostatistics
Keith Worsley,
McGill
Jonathan Taylor,
Stanford and Université de Montréal
Arnaud Charil,
Montreal Neurological Institute
CfA red shift survey, FWHM=13.3
100
80
Euler Characteristic (EC)
60
"Meat ball"
topology
40
20
"Bubble"
topology
0
-20
-40
"Sponge"
topology
-60
-80
-100
-5
CfA
Random
Expected
-4
-3
-2
-1
0
1
Gaussian threshold
2
3
4
5
Brain imaging
Detect sparse regions of “activation”
Construct a test statistic image for detecting activation
Activated regions: test statistic > threshold
Choose threshold to control false positive rate to say 0.05
i.e. P(max test statistic > threshold) = 0.05
Bonferroni???
Â
¹=
max Z1 cos µ + Z2 sin µ
0·µ·¼=2
Example test statistic:
Z1~N(0,1)
Z2~N(0,1)
s2
3
2
1
0
-1
-2
Excursion sets,
Xt = fs : Â
¹ ¸ tg s1
-3
Threshold
t 4
Rejection regions,
Z2
2
Search
Region,
S
Rt = fZ : Â
¹ ¸ tg
3
2
1
0
Z1
-2
-2
0
2
Euler characteristic heuristic
Search Region, S
Euler characteristic, EC
EC= 1
7
Excursion sets, Xt
6
5
2
1
1
0
10
Observed
8
¸ t)
P(max Â(s)
¹
s2S
6
¼ E(EC) = 0:05
Expected
) t = 3:75
4
2
0
-2
0
0.5
1
1.5
E(EC(S \ Xt )) =
2
X
D
d=0
2.5
L (S)½ (t)
d
d
3
3.5
4
Threshold, t
E(EC(S \ Xt )) =
X
D
d=0
Tube(λS,r)
Radius, r
Tube(Rt,r)
14
L (S)½ (t)
d
d
r
12
10
µ
@Z
@s
¸ = Sd
p
4 log 2
=
FWHM
¶
r
0
1
4
0.5
2
Z1
0.2
Area
0
jTube(¸S; r)j =
X
D
d=0
2L1 (S)r
¼ L0 (S)r2
0
0.5
1
1.5
2
Radius of Tube, r
¼d
L
¡d (S)r d
D
¡(d=2 + 1)
Probability
L (S)
2
50
0
EC
½d (t)
-2 density
0
2
0.4
jTube(¸S; r)j
100
0.4
-2
L (S)0
Lipschitz-Killing
2 4 6 curvature
8 10 12 14
d
150
0.8
0.6
λS
6
Rt
2
1.5
8
1
Z2
2
Radius, r
P(Tube(Rt ; r))
0.3
p
2¼½1 (t)r
½0 (t)
¼½2 (t)r2
0.2
0.1
0
0
0.5
Radius of
1 Tube, r
1
X
P(Tube(Rt ; r)) =
(2¼)d=2 ½d (t)rd =d!
d=0
EC density ½d (t)
of the Â
¹ statistic
Z2~N(0,1)
Tube(Rt,r)
r
Rejection region
Rt
t-r t
Z1~N(0,1)
Taylor’s Gaussian Kinematic Formula:
1
X
P (Z1 ; Z2 2 Tube(Rt ; r)) =
(2¼)d=2 ½d (t)rd =d!
d=0
½0 (t) =
Z
= ½0 (t) + (2¼)1=2 ½1 (t)r + (2¼)½2 (t)r2 =2 + ¢ ¢ ¢
Z 1
=
(2¼)¡1=2 e¡z2 =2 dz + e¡(t¡r)2 =2 =4
t¡r
1
(2¼)¡1=2 e¡z2 =2 dz + e¡t2 =2 =4
t
½1 (t) = (2¼)¡1 e¡t2 =2 + (2¼)¡1=2 e¡t2 =2 t=4
½2 (t) = (2¼)¡3=2 e¡t2 =2 t + (2¼)¡1 e¡t2 =2 (t2 ¡ 1)=8
..
.
Lipschitz-Killing
curvature Ld (S)
r
Tube(λS,r)
λS
Steiner-Weyl Volume of Tubes Formula:
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)
= Resels0 (S)
0
p
L (S) = 1 Perimeter(¸S) = 4 log 2 Resels (S)
1
1
2
L (S) = Area(¸S)
= 4 log 2 Resels2 (S)
2
How to ¯nd Lipschitz-Killing curvature Ld (S)
Edge length × λ
FWHM/√(4log2)
12
10
8
6
4
2
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10
of simplices
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 simplices
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
( )
Non-isotropic data?
µ
¸ = Sd
Z~N(0,1)
s2
3
@Z
@s
¶
p
=
4 log 2
FWHM
2
1
0.14
0.12
0
-1
-2
s1
12
10
8
6
4
2
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6
-3
Can we warp the data to isotropy?
.. .
i.e. multiply edge lengths by λ?
. . ..
. . . .
. . . ...
Locally
. . . . . yes, but we may need extra dimensions.
. . . . .
. . . .
. . . . ..... Nash Embedding Theorem:
. . . . .
dimensions ≤ D + D(D+1)/2
. . . ....
. . ...
. . .
D=2: dimensions ≤ 5
8
10
0.1
0.08
0.06
Warping to isotropy not needed – only warp
the ptriangles
µ
¶
¸ = Sd
Z~N(0,1)
s2
3
@Z
@s
=
4 log 2
FWHM
2
1
0.14
0.12
0
-1
-2
Edge length × λ
FWHM/√(4log2)
12
10
8
6
4
2
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10
0.1
0.08
0.06
-3
s1
of simplices
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 simplices
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
Z5
Z6
Z7
Z9 … Zn
Z8
Replace coordinates of the simplices in S⊂RealD by
(Z1,…,Zn) / ||(Z1,…,Zn)|| in Realn
of simplices
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 simplices
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
Unbiased!
( )
Unbiased!
MS lesions and cortical
thickness





Idea: MS lesions interrupt neuronal signals,
causing thinning in down-stream cortex
Data: n = 425 mild MS patients
Lesion density, smoothed 10mm
Cortical thickness, smoothed 20mm
Find connectivity i.e. find voxels in 3D, nodes
in 2D with high


correlation(lesion density, cortical thickness)
Look for high negative correlations …
n=425 subjects, correlation = -0.568
Average cortical thickness
5.5
5
4.5
4
3.5
3
2.5
2
1.5
0
10
20
30
40
50
60
Average lesion volume
70
80
Thresholding? Cross
correlation random field

Correlation between 2 fields at 2 different
locations, searched over all pairs of locations


one in R (D dimensions), one in S (E dimensions)
sample size n
Cao & Worsley, Annals of Applied Probability (1999)

MS lesion data: P=0.05, c=0.325, T=7.07
Normalization


LD=lesion density, CT=cortical thickness
Simple correlation:


Subtracting global mean thickness:


Cor( LD, CT )
Cor( LD, CT – avsurf(CT) )
And removing overall lesion effect:

Cor( LD – avWM(LD), CT – avsurf(CT) )
0.1
correlation
0
5
x 10
2.5
2
-0.1
1.5
-0.2
1
-0.3
threshold
-0.4
-0.5
0
50
100
150
Different hemisphere
0.1
5
x 10
2.5
0
correlation
Histogram Same hemisphere
-0.1
2
-0.2
1.5
-0.3
1
0.5
-0.4
0
-0.5
0
threshold
50
100
150
0.5
0
‘Conditional’ histogram: scaled to same max at each distance
0.1
1
-0.1
0.6
-0.2
0.4
-0.3
-0.4
-0.5
0
threshold
50
100 150
distance (mm)
1
0
0.8
correlation
correlation
0
0.1
0.8
-0.1
0.6
-0.2
0.4
-0.3
0.2
-0.4
0
-0.5
0
threshold
50
100 150
distance (mm)
0.2
0