bayesian probabilistic or weights of evidence model for mineral prospectivity mapping
DESCRIPTION
Bayesian Probabilistic or Weights of Evidence Model for Mineral Prospectivity Mapping. Probabilistic model (Weights of Evidence). What is needed for the WofE calculations? A training point layer – i.e. known mineral deposits; One or more predictor maps in raster format. - PowerPoint PPT PresentationTRANSCRIPT
Indian Institute of Technology Bombay
Bayesian Probabilistic or Weights of Evidence Model for Mineral Prospectivity Mapping
Indian Institute of Technology Bombay
2
Probabilistic model (Weights of Evidence)
• What is needed for the WofE calculations?– A training point layer –
i.e. known mineral deposits;
– One or more predictor maps in raster format.
Indian Institute of Technology Bombay
PROBABILISTIC MODELS (Weights of Evidence or WofE)
Four steps:1. Convert multiclass maps to binary maps2. Calculation of prior probability3. Calculate weights of evidence (conditional probability) for
each predictor map4. Combine weights
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• The probability of the occurrence of the targeted mineral deposit type when no other geological information about the area is available or considered.
Study area (S)
Target deposits D
Assuming- 1. Unit cell size = 1 sq km2. Each deposit occupies 1 unit cell
Total study area = Area (S) = 10 km x 10 km = 100 sq km = 100 unit cells
Area where deposits are present = Area (D) = 10 unit cells
Prior Probability of occurrence of deposits = P {D} = Area(D)/Area(S)= 10/100 = 0.1
Prior odds of occurrence of deposits = P{D}/(1-P{D}) = 0.1/0.9 = 0.1110k
10k
1k1k
Calculation of Prior Probability
Indian Institute of Technology Bombay
5
Convert multiclass maps into binary maps
• Define a threshold value, use the threshold for reclassification
Multiclass map
Binary map
Indian Institute of Technology Bombay • How do we define the threshold?
Use the distance at which there is maximum spatial association as the threshold !
Convert multiclass maps into binary maps
Indian Institute of Technology Bombay
• Spatial association – spatial correlation of deposit locations with geological feature.
A
BC
D
A
BC
D
10km
10km
1km
1km
Gold Deposit (D) Study area (S)
Convert multiclass maps into binary maps
Indian Institute of Technology Bombay
A
B C
D
Which polygon has the highest spatial association with D?More importantly, does any polygon has a positive spatial association with D ???
What is the expected distribution of deposits in each polygon, assuming that they were randomly distributed? What is the observed distribution of deposits in each polygon?
Positive spatial association – more deposits in a polygon than you would expect if the deposits were randomly distributed.
If observed >> expected; positive associationIf observed = expected; no association If observed << expected; negative association
Convert multiclass maps into binary maps
Indian Institute of Technology Bombay
A
B C
D
Area (A) = n(A) = 25; n(D|A) = 2Area (B) = n(A) = 21; n(D|B) = 2Area(C) = n(C) = 7; n(D|C) = 2 Area(D) = n(D) = 47; n(D|D) = 4Area (S) = n(S) = 100; n(D) = 10
OBSERVED DISTRIBUTION
Convert multiclass maps into binary maps
Indian Institute of Technology Bombay
A
B C
DArea (A) = n(A) = 25; n(D|A) = 2.5Area (B) = n(A) = 21; n(D|B) = 2.1Area(C) = n(C) = 7; n(D|C) = 0.7Area(D) = n(D) = 47; n(D|D) = 4.7(Area (S) = n(S) = 100; n(D) = 10)
EXPECTED DISTRIBUTION
Expected number of deposits in A = (Area (A)/Area(S))*Total number of deposits
Convert multiclass maps into binary maps
Indian Institute of Technology BombayA
B C
D
Area (A) = n(A) = 25; n(D|A) = 2.5
Area (B) = n(A) = 21; n(D|B) = 2.1
Area(C) = n(C) = 7; n(D|C) = 0.7
Area(D) = n(D) = 47; n(D|D) = 4.7
(Area (S) = n(S) = 100; n(D) = 10)
EXPECTED DISTRIBUTIONArea (A) = n(A) = 25; n(D|A) = 2
Area (B) = n(A) = 21; n(D|B) = 2
Area(C) = n(C) = 7; n(D|C) = 2
Area(D) = n(D) = 47; n(D|D) = 4
Area (S) = n(S) = 100; n(D) = 10
OBSERVED DISTRIBUTION
Only C has positive association!So, A, B and D are classified as 0; C is classified as 1.
Another way of calculating the spatial association : = Observed proportion of deposits/ Expected proportion of deposits= Proportion of deposits in the polygon/Proportion of the area of the polygon= [n(D|A)/n(D)]/[n(A)/n(S)] • Positive if this ratio >1• Nil if this ratio = 1 • Negative if this ratio is < 1
Convert multiclass maps into binary maps
Indian Institute of Technology Bombay
LA
BC
D
10km
10km
1km
1km
Gold Deposit (D) Study area (S)
Convert multiclass maps into binary maps – Line features
Indian Institute of Technology Bombay
1km
1km
Gold Deposit (D)
10
0
000
0
000
11
112
11
1
1111
111
111
1
11
2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 2 3 4 5 6 7 8
2 3 4 5 6 7 2 3 4 5 6 2 3 4 5 6
2 3 4 5 2 3 4 2 3 4 2 3 4
3 2 3 2 4 3 2 4 3 2
5 4 3 2 5 4 3 2
Distance from
the fault
No. of
pixels
No of deposit
s
Ratio (Observe
d to Expected
)0 9 1 1.11 21 3 1.42 17 0 0.03 16 3 1.94 14 2 1.45 9 0 0.06 6 0 0.07 4 0 0.08 3 1 3.39 1 0 0.0
Convert multiclass maps into binary maps – Line features
Indian Institute of Technology Bombay
• Calculate observed vs expected distribution of deposits for cumulative distances
Gold Deposit (D)
10
0
000
0
000
11
112
11
1
1111
111
111
1
11
2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 2 3 4 5 6 7 8
2 3 4 5 6 7 2 3 4 5 6 2 3 4 5 6
2 3 4 5 2 3 4 2 3 4 2 3 4
3 2 3 2 4 3 2 4 3 2
5 4 3 2 5 4 3 2
Distance from the fault
No. of pixels
Cumul No. of pixels
No of deposits
Cumul No. of deposits
Ratio (Observed to Expected)
0 9 9 1 1 1.1
1 21 30 3 4 1.3
2 17 47 0 4 0.9
3 16 63 3 7 1.1
4 14 77 2 9 1.2
5 9 86 0 9 1.0
6 6 92 0 9 1.0
7 4 96 0 9 0.9
8 3 99 1 10 1.0
9 1 100 0 10 1.0
=< 1 – positive association (Reclassified into 1)>1 – negative association (Reclassified into 0)
Convert multiclass maps into binary maps – Line features
Indian Institute of Technology Bombay
Weights of evidence ~ quantified spatial associations of deposits with geological features
Study area (S)
10k
Target deposits10k
Unit cell
1k1k
Objective: To estimate the probability of occurrence of D in each unit cell of the study area
Approach: Use BAYES’ THEOREM for updating the prior probability of the occurrence of mineral deposit to posterior probability based on the conditional probabilities (or weights of evidence) of the geological features.
Calculation of Weights of Evidence
Geological Feature (B1) Geological Feature (B2)
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P{D|B} = P{D& B}
P{B}= P{D} P{B|D}
P{B}
P{D|B} = P{D & B}
P{B}= P{D} P{B|D}
P{B}
Posterior probability of D given the presence of B
Posterior probability of D given the absence of B
Bayes’ theorem:D- Deposit
B- Geol. Feature
THE BAYES EQUATION ESTIMATES THE PROBABILTY OF A DEPOSIT GIVEN THE
GEOLOGICAL FEATURE FROM THE PROBABILITY OF THE FEATURE GIVEN THE DEPOSITS
ObservationInference
Calculation of Weights of Evidence
Indian Institute of Technology Bombay
It has been observed that on an average 100 gold deposits occur in 10,000 sq km area of specific geological areas. In such areas, 80% of deposits occur in Ultramafic (UM) rocks, however, 9.6% of barren areas also occur in Ultramafic rocks. You are exploring a 1 sq km area of an Archaean geological province with Ultramafic rocks (UM). What is the probability that the area will contain a gold deposit? Assume that a gold deposit occupies 1 sq km area.
EXERCISE
P(D|UM) = P(D) x [P(UM|D) / P(UM)]
P(D) = n(D)/n(S)
P(UM|D) = n(UM & D)/n(D)
P(UM) = n(UM)/n(S)
P(D|UM) =
(100/10000) * [(80/100)/(1030.4/10000)]
= 0.077
n(S) =
n(D) =
n(UM&D) =
n(UM) =
n(UM) =
10,000100
80?80% of 100 + 9.6% (10,000 - 100)
= 1030.4
Indian Institute of Technology Bombay
Using odds (P/(1-P)) formulation:
O{D|B} = O{D} P{B|D}
P{B|D}Odds of D given the presence of B
O{D|B} = O{D} P{B|D}
P{B|D}Odds of D given the absence of B
Taking logs on both sides:
Loge (O{D|B}) = Loge(O{D}) + Log of odds of D given the presence of BP{B|D}P{B|D}
loge
Loge (O{D|B}) = Loge(O{D}) + Log of odds of D given the absence of BP{B|D}P{B|D}
loge
+ive weight of evidence (W+)
-ive weight of evidence (W-)
Calculation of Weights of Evidence
Indian Institute of Technology Bombay Contrast (C) measures the net strength of spatial association between the
geological feature and mineral deposits
Contrast = W+ – W-
+ ive Contrast – net positive spatial association
-ive Contrast – net negative spatial association
zero Contrast – no spatial association
Can be used to test spatial associations
Calculation of contrast
Indian Institute of Technology Bombay
Total number of cells in study area: n(S)Total number of cells occupied by deposits (D): n(D)Total number of cells occupied by the feature (B): n(B)Total number of cells occupied by both feature and deposit: n(B&D)
= n( )/n(D) = n( )/ = n( )/n(D) = n( )/
B & D
B & DP{B|D}P{B|D} B & D
P{B|D}
P{B|D} B & D
n(D)
n(D)
B1
D
B2
B1 D B1 D
B1 D
B1 D
B1
SD
Calculation of Probabilty
P(D) = n(D)/n(S)
Indian Institute of Technology Bombay
Basic quantities for estimating weights of evidenceTotal number of cells in study area: n(S)Total number of cells occupied by deposits (D): n(D)Total number of cells occupied by the feature (B): n(B)Total number of cells occupied by both feature and deposit: n(B&D)
Derivative quantities for estimating weights of evidenceTotal number of cells not occupied by D: n( ) = n(S) – n(D) Total number of cells not occupied by B: n( ) = n(S) – n(B)Total number of cells occupied by B but not D: n( B & D) = n(B) – n( B & D)Total number of cells occupied by D but not B: n(B & D) = n(D) – n(B & D)Total number of cells occupied by neither D but nor B: n( B & D) = n(S) – n(B) – n(D) + n( B & D)
DB
Probabilities are estimated as area (or no. of unit cells) proportions
P{B|D}P{B|D}
loge
W+ = P{B|D}P{B|D}
loge
W- =
= n( )/n(D) = n( )/ = n( )/n(D) = n( )/
B & D
B & DP{B|D}P{B|D} B & D
P{B|D}
P{B|D} B & D
n(D)
n(D)
Where,
B1
D
B2
B2 D B2 D
B2 D
B2 D
B1 D B1 D
B1 D
B1 D
B1
SD
B2
SD
Calculation of Weights of Evidence
Indian Institute of Technology Bombay
Exercise
B2 D B2 D
B2 D
B2 DB1 D B1 D
B1 D
B1 D
10k
10kB
1
B2
S
B1
SD B
2S
D
Unit cell size = 1 sq km & each deposit occupies 1 unit cell
n(S) = 100n(D) = 10n(B1) = 16n(B2) = 25n(B1 & D) = 4n(B2 & D) = 3
Calculate the weights of evidence (W+ and W-) and Contrast values for B1 and B2
= n( )/n(D) = n( )/ = [n(B) – n( )]/[n(S) –n(D)] = n( )/n(D) = [n(D) – n( )]/n(D) = n( )/ = [n(S) – n(B) – n(D) + n( )]/[n(S) –
n(D)]
B & D
B & DP{B|D}P{B|D} B & D
P{B|D}
P{B|D} B & D
n(D)
n(D)
Where,B &
D
B & D
B & D
P{B|D}P{B|D}
loge
W+ = P{B|D}P{B|D}
loge
W- =
Indian Institute of Technology Bombay
Loge (O{D|B}) = Loge(O{D}) + W+B
Loge (O{D|B}) = Loge(O{D}) + W-B
Assuming conditional independence of
the geological features B1 and B2, the
posterior probability of D given B1 and B2
can be estimated using:
Loge (O{D|B1, B2}) = Loge(O{D}) + W+B1 + W+B2
Loge (O{D|B1, B2}) = Loge(O{D}) + W-B1 + W+B2
Loge (O{D|B1, B2}) = Loge(O{D}) + W+B1 + W-B2
Loge (O{D|B1, B2}) =
Loge(O{D}) + W-B1 + W-B2
Probability of D given the presence of B1 and B2
Probability of D given the absence of B1 and presence B2
Probability of D given the presence of B1 and absence B2
Probability of D given the absence of B1 and B2
Loge (O{D|B1, B2, … Bn}) = Loge(O{D}) + ∑W+/-Bii=1
nOr in general, for n geological features,
The sign of W is +ive or -ive, depending on whether the feature is absent or present
The odds are converted back to posterior probability using the relation 0 = P/(1+P)
Combining Weights of Evidence: Posterior Probability
Feature B2 Feature B1
Deposit D
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Loge (O{D|B1, B2}) =
Loge(O{D}) + ∑W+/-Bii=1
n
Calculation of posterior probability (or odds) require:• Calculation of pr prob (or odds) of occurrence of deposits in the study area• Calculation of weights of evidence of all geological features, i.e,
P{B|D}P{B|D}
loge
P{B|D}P{B|D}
loge
W+ =
W- =
&
Combining Weights of Evidence: Posterior Probability
Indian Institute of Technology Bombay Loge (O{D|B1, B2})
= Loge(O{D}) + W+/-B1 +
W+/-B2 Loge(O{D}) = Loge(0.11) = -
2.2073
Calculate posterior probability given:
1. Presence of B1 and B2;2. Presence of B1 and absence of
B2;3. Absence of B1 and presence of
B2;4. Absence of both B1 and B2
B1
B2
S
Prior Prb = 0.10Prior Odds =
0.11
Combining Weights of Evidence: Posterior Probability
Indian Institute of Technology Bombay
Loge (O{D|B1, B2}) =
Loge(O{D}) + W+/-B1 + W+/-B2
Loge (O{D|B1, B2}) = -2.2073 + 1.0988 + 0.2050 = -0.8585
O{D|B1, B2} = Antiloge (-0.8585) = 0.4238 P = O/(1+O) = (0.4238)/(1.4238) = 0.2968
For the areas where both B1 and B2 are present
Loge (O{D|B1, B2}) = -2.2073 + 1.0988 - 0.0763 = -1.1848
O{D|B1, B2} = Antiloge (- 1.1848) = 0.3058 P = O/(1+O) = (0.3058)/(1.3058) = 0.2342
For the areas where B1 is present but B2 is absent
Loge (O{D|B1, B2}) = -2.2073 - 0.3678 + 0.2050 = -2.3701
O{D|B1, B2} = Antiloge (-2.3701) = 0.0934 P = O/(1+O) = (0.0934)/(1.0934) = 0.0854
Loge (O{D|B1, B2}) = -2.2073 - 0.3678 - 0.0763 = -2.6514
O{D|B1, B2} = Antiloge (-2.6514) = 0.0705 P = O/(1+O) = (0.0705)/(1.0705) = 0.0658
For the areas where both B1 and B2 are absent
For the areas where B1 is absent but B2 is present
Loge(O{D}) = Loge(0.11) = -2.2073
Posterior probability0.29680.2342
0.08540.0658
Prospectivity Map
B1
B2
S
Prior Prb = 0.10
Combining Weights of Evidence: Posterior Probability