Multi-type patterns
Marked point processes: Z (si ) observed at event locations
e.g. height or diameter of a tree
plant died / alive
Z only exists at event locations
Multivariate point processes: subset of marked processes
marks are a small number of discrete values
live/dead, species of plant, fatality / not at an accident location
often 2 types
concepts often extend to > 2 types
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
1 / 27
Questions that might be asked
1) Single type questions asked separately for each type
we know how to do this, will not discuss
2) do two types tend to occur together (i.e. in same areas)
Two species of tupelo in the swamp data: NX is Nyssa aquatica, NS is
Nyssa sylvatica. Do they tend to occur in same areas?
If they do, then two processes: Nyssa locations, then flip a coin to
determine species
λNS (s)
=
PNS λ(s)
λNX (s)
=
(1 − PNS )λ(s)
Two-stage mechanism often makes a lot of sense, e.g. tree mortality
A spatial process determines where trees are located
A second, possibly spatial process, determines whether live or dead
If they don’t occur together, have spatial segregation
Maybe even clusters of NX and clusters of NS
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
2 / 27
Spatial epidemiology
Understand spatial aspects of disease
In particular, are disease cases clustered?
Answer provides clues to disease mechanism
Data are locations of events (disease cases)
Can not just look at clustering of cases
Often population is clustered
So if P[disease] is constant, not spatially varying
still expect to find clustering of cases
Get random sample of controls (not diseased) from the general
population
Two types of events: case or control, ask questions like:
Are cases more clustered than the general population?
Is relative risk of disease = λcase /λcontrol constant
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
3 / 27
June 2013
4 / 27
Pictures: spatial segregation
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Spatial Data Analysis - Part 12
June 2013
5 / 27
June 2013
6 / 27
Pictures: equally clustered?
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Spatial Data Analysis - Part 12
Multi-type K function
extend definition of K(x) to multiple types
“cross-K”: KAB (x) = λ1B E #B events w/i x of an A
note: KAB (x) = KBA (x), although estimates not identical because of
edge effects
“self-K”: KAA (x) = λ1A E #A events w/i x of an A
look at KAA (x) − KAB (x).
If > 0, spatial segregation. A’s more common around A’s
If < 0, spatial association. B’s more common around A’s
minor detail if more than 2 types:
compare KAA to what?
usual answer is KA. , i.e. A to any type
KAA − KA. gives same answers as KAA − KA,−A
Test significance of departure from 0 by random labeling
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
7 / 27
June 2013
8 / 27
Random labeling
Fix the locations of events
Randomly reassign labels (type of event)
Compute statistic of interest
pointwise difference in K functions
some summary statistic
Corresponds to the “two processes” view
one process generating locations
a second process generating labels (types)
is that second process a random labelling of locations?
Makes a lot of sense for live/dead case/control labels
less clear for multiple species
Alternatives exist, talk about later
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
Why does this work?
Why does comparing KAA (x) to KAB (x) work?
Two processes view
locations all have a Kall (x) function.
imagine a random sample of locations: A, e.g. τ = 0.4
what is the relationship between Kall (x) and KA (x)?
intensity: λA = τ λall = 0.4λall
expected count of A within r of an A: E nA = τ expected count of all
within r of an A
so KA (x) = EλAnA = τ τEλallnall = Eλallnall = Kall (x)
Randomly “thinning” a process does not change the K function
So random sample of controls gives K for the population
If cases are also a random sample of the population,
Kcase (x) = Kcontrol (x)
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
9 / 27
Nyssa species in the Savannah River Swamp
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c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
10 / 27
1.0
^
K ss K i so (r ) π − r
0.5
^ i so
K sa K N S, N S(r ) π − r
^ i so
K as K N S, N S(r ) π − r
^
K aa K i so (r ) π − r
−0.5
0.0
K NS, NS (r ) π − r
1.5
Nyssa L(x) functions
0
2
c Philip M. Dixon (Iowa State Univ.)
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r
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10
Spatial Data Analysis - Part 12
12
June 2013
11 / 27
150
More NS than NA around NS
0
.
50
100
Kss − Ksa
0
c Philip M. Dixon (Iowa State Univ.)
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Spatial Data Analysis - Part 12
8
10
12
June 2013
12 / 27
More NA than NS around NA
−50
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.
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100
Kaa − Kas
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c Philip M. Dixon (Iowa State Univ.)
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Spatial Data Analysis - Part 12
12
June 2013
13 / 27
More questions
4) Is one type of event more clustered than the other
i.e. for short distances, is KAA (x) < KBB (x)?
approach in same way as 3)
compare observed difference K̂AA (x) − K̂BB (x) to a simulation envelope
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
14 / 27
1.0
^
K ss K i so (r ) π − r
0.5
^ i so
K sa K N S, N S(r ) π − r
^ i so
K as K N S, N S(r ) π − r
^
K aa K i so (r ) π − r
−0.5
0.0
K NS, NS (r ) π − r
1.5
Nyssa L(x) functions
0
2
c Philip M. Dixon (Iowa State Univ.)
4
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Spatial Data Analysis - Part 12
12
June 2013
15 / 27
80
Are NS more clustered than NA?
−60
−20
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.
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40
60
Kss − Kaa
0
c Philip M. Dixon (Iowa State Univ.)
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Spatial Data Analysis - Part 12
8
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June 2013
16 / 27
More questions
5) Are two processes independent?
Not exactly same question as 3), random labelling.
Not viewed as two stages: locations, then labels
Instead, is P[event B in area dA] independent of presence of A’s?
P[event B in area dA] may depend on presence of other B’s
Turns out to be difficult to evaluate
want to maintain all characteristics of the process for A and all the
characteristics of the process for B, but break any dependence.
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
17 / 27
Testing independence
Much harder than testing random labeling
Have two processes: A and B.
Each has whatever structure it has
Question is whether A pattern is placed down independently of the B
pattern
Need to maintain the A pattern, maintain the B pattern, but break
any “connection” between them
Historical approach: toroidal rotations
randomly displace A horizontally and vertically
Maintain all features of process A and process B, but allows A to shift
vis-a-vis locations of B = independence
points that shift “out” of the study area reappear on other side (next
slide)
Again, compare observed difference to simulated differences
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
18 / 27
June 2013
19 / 27
Testing independence
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c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
Testing independence
Issues / concerns:
only feasible for rectangular study areas
Now understood to be very sensitive to assumption of first order
stationarity (constant intensity across study area)
If intensity varies, gives very misleading results
New approach: simulate realizations of A that match specified
properties of observed A
Ditto B, both independent
e.g. K function, intensity over the study area, perhaps others too
Tscheschel and Stoyan, 2006, Comp. Stats. and Data Analysis
51:859-871.
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
20 / 27
Intensity and relative risk
Intensity: different insights
Can estimate intensity for each type separately (next slide)
choose bandwidth separately for each type
Or intensity for events, ignoring type
Sum of the type-specific intensity
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
21 / 27
Intensity and relative risk
0.02
0.06
N. sylvatica
0.02
0.06
N. aquatica
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
22 / 27
Intensity and relative risk
In many problems, it makes sense to think about the risk of one type
of event
λA
λA
r=
=
λ
λA + λB
“risk” is the probability that a randomly chosen event is a type A
event
Allow everything to vary over the study area
each intensity is
λA (s) = r (s) λ(s)
λB (s) = (1 − r (s)) λ(s)
Can adjust one or both if a random sample, not an enumeration
Note:
λA (s)
r (s)
=
λB (s)
1 − r (s)
the odds ratio, super important in epidemiology
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
23 / 27
June 2013
24 / 27
Intensity and relative risk
Estimating r (s)
model λA (s) and λB (s) as functions of covariates
model r (s) directly as function of covariates
or non-parametric (smoothing) estimate of r (s)
NP smoothing: based on λ̂A (s) and λ̂B (s)
use same bandwidth for both components
Usual estimator: based on point-process likelihood
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
Intensity and relative risk
cv (σ)
0.60
0.65
0.70
0.75
0.80
BW for relative risk
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14
σ
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
25 / 27
Intensity and relative risk
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c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12
June 2013
26 / 27
June 2013
27 / 27
Summary of multi-type point pattern analysis
Lots of things you could do
What question(s) is/are most important?
Focus the analysis on answering those questions
What if the marks are continuous (not just a few types)
mark correlation function
describe correlation between marks on two events
and how those change with distance between events
few examples, not well understood
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 12