Indian Institute of Technology Bombay Bayesian Probabilistic or Weights of Evidence Model for Mineral Prospectivity Mapping

Indian Institute of Technology Bombay 2 Bayes Rule Probability of an event happening given an observation can be estimated by probability of that observation given the event. P{D|B} = P{D& B} P{B} = P{D} P{B|D} P{B} Posterior probability of D given the presence of B Bayes’ theorem: D- Event B- Observation Observation Inference

Indian Institute of Technology Bombay 3 A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. You are given the following data: - 15% of the cabs in the city are Green and 85% are Blue. -A witness identified the cab as Blue. The court tested the reliability of the witness under the same circumstances that existed on the night of the accident and concluded that the witness correctly identified each one of the two colours 80% of the time and failed 20% of the time. What is the probability that the cab involved in the accident was Blue rather than Green? Which cab was involved in the accident?

Indian Institute of Technology Bombay 4 Which cab was involved in the accident? P(G) =.15 This is the base rate of green cabs in the city. It gives the prior probability that the cab in the accident is green. Similarly, P(B) =.85. P(SG|G) =.80. This is the probability the witness will be correct in saying green when the cab in fact is green, i.e., given that the cab really was green. Similarly, P(SB|B) =.80. These are the probabilities that witnesses are correct, so by the complement rule, the probabilities of misidentifications are: P(SG|B) =.2 and P(SB|G) =.2. What we want to know is the probability that the cab really was green, given that Wilbur said it was green, i.e., we want to know P(G|SG). According to Bayes’ rule, this probability is given by: P(G|SG) =P(G)×Pr(SG|G) / Pr(SG). We have the values for the two expressions in the numerator: P(G) =.15 and P(SG|G) =.8, but we have to do a little work to determine the value for the expression P(SG) in the denominator. According to the total probability rule: P(SG) = P(G)×P(SG|G)+P(B)×P(SG|B) = (.15×.80)+(.85×.20) = =.29 Finally, we substitute this number,.29, into the denominator of Bayes’ Rule: P(G|SG) = P(G)×P(SG|G) / P(SG) = 15×.80 /.29 =.414

Indian Institute of Technology Bombay 5 Probabilistic model (Weights of Evidence) What is needed for the WofE calculations? training point –A training point layer – i.e. known mineral deposits; predictor maps raster –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 maps 2.Calculation of prior probability 3.Calculate weights of evidence (conditional probability) for each predictor map 4. Combine weights

Indian Institute of Technology Bombay 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 km 2.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 = k 1k Calculation of Prior Probability

Indian Institute of Technology Bombay 8 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 B C D A B C D 10km 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 association If 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) = 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 Convert multiclass maps into binary maps

Indian Institute of Technology Bombay A 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 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 Bombay A 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 DISTRIBUTION Area (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 L A B C D 10km 1km Gold Deposit (D) Study area (S) Convert multiclass maps into binary maps – Line features

Indian Institute of Technology Bombay 1km Gold Deposit (D) Distance from the fault No. of pixels No of deposits Ratio (Observed to Expected) 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) Distance from the fault No. of pixels Cumul No. of pixels No of deposits Cumul No. of deposits Ratio (Observed to Expected) =< 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 deposits 10k Unit cell 1k 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)

Indian Institute of Technology Bombay 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 Observation Inference 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)] = n(S) = n(D) = n(UM&D) = n(UM) = 10, ? 80% of % (10, ) =

Indian Institute of Technology Bombay Using odds (P/(1-P)) formulation: O{D|B} = O{D} P{B|D} Odds of D given the presence of B O{D|B} = O{D} P{B|D} Odds of D given the absence of B Taking logs on both sides: Log e (O{D|B}) = Log e (O{D}) + Log of odds of D given the presence of B P{B|D} log e Log e (O{D|B}) = Log e (O{D}) + Log of odds of D given the absence of B P{B|D} log e +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 P{B|D} B & D P{B|D} B & D n(D) B1 D B2 B1 D B1 S D Calculation of Probabilty P(D) = n(D)/n(S)

Indian Institute of Technology Bombay Basic quantities for estimating weights of evidence 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) Derivative quantities for estimating weights of evidence Total 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) D B Probabilities are estimated as area (or no. of unit cells) proportions P{B|D} log e W+ = P{B|D} log e W- = = n( )/n(D) = n( )/ = n( )/n(D) = n( )/ B & D P{B|D} B & D P{B|D} B & D n(D) Where, B1 D B2 B2 D B1 D B1 S D B2 S D Calculation of Weights of Evidence

Indian Institute of Technology Bombay Exercise B2 D B1 D 10k B1 B2 S B1 S D B2 S D Unit cell size = 1 sq km & each deposit occupies 1 unit cell n(S) = 100 n(D) = 10 n(B1) = 16 n(B2) = 25 n(B1 & D) = 4 n(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} B & D P{B|D} B & D n(D) Where, B & D P{B|D} log e W+ = P{B|D} log e W- =

Indian Institute of Technology Bombay Log e (O{D|B}) = Log e (O{D}) + W+ B Log e (O{D|B}) = Log e (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: Log e (O{D|B1, B2}) = Log e (O{D}) + W+ B1 + W+ B2 Log e (O{D|B1, B2}) = Log e (O{D}) + W- B1 + W+ B2 Log e (O{D|B1, B2}) = Log e (O{D}) + W+ B1 + W- B2 Log e (O{D|B1, B2}) = Log e (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 Log e (O{D|B1, B2, … Bn}) = Log e (O{D}) + ∑W+/- Bi i=1 n Or 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

Indian Institute of Technology Bombay Log e (O{D|B1, B2}) = Log e (O{D}) + ∑W+/- Bi i=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} log e P{B|D} log e W+ = W- = & Combining Weights of Evidence: Posterior Probability

Indian Institute of Technology Bombay Log e (O{D|B1, B2}) = Log e (O{D}) + W+/- B1 + W+/- B2 Log e (O{D}) = Log e (0.11) = 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 = 0.10 Prior Odds = 0.11 Combining Weights of Evidence: Posterior Probability

Indian Institute of Technology Bombay Log e (O{D|B1, B2}) = Log e (O{D}) + W+/- B1 + W+/- B2 Log e (O{D|B1, B2}) = = O{D|B1, B2} = Antilog e ( ) = P = O/(1+O) = (0.4238)/(1.4238) = For the areas where both B1 and B2 are present Log e (O{D|B1, B2}) = = O{D|B1, B2} = Antilog e ( ) = P = O/(1+O) = (0.3058)/(1.3058) = For the areas where B1 is present but B2 is absent Log e (O{D|B1, B2}) = = O{D|B1, B2} = Antilog e ( ) = P = O/(1+O) = (0.0934)/(1.0934) = Log e (O{D|B1, B2}) = = O{D|B1, B2} = Antilog e ( ) = P = O/(1+O) = (0.0705)/(1.0705) = For the areas where both B1 and B2 are absent For the areas where B1 is absent but B2 is present Log e (O{D}) = Log e (0.11) = Posterior probability Prospectivity Map B1 B2 S Prior Prb = 0.10 Combining Weights of Evidence: Posterior Probability