bayesian probabilistic or weights of evidence model for mineral prospectivity mapping

Indian Institute of Technology Bombay

Bayesian Probabilistic or Weights of Evidence Model for Mineral Prospectivity Mapping


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.


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


• 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


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 !



• 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)



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



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



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


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



LA

BC

D

10km

10km

1km

1km

Gold Deposit (D) Study area (S)

Convert multiclass maps into binary maps – Line features


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



• 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)



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)


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



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


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-)


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


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)


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


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



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


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- =


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



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- =

&


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




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


Loge (O{D|B1, B2}) = -2.2073 - 0.3678 - 0.0763 = -2.6514


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


bayesian probabilistic or weights of evidence model for mineral prospectivity mapping

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