Chapter 5 - Problem 8 - Bayes' Theorem Example

# Chapter 5 - Problem 8 - Bayes' Theorem Example - Chapter 5...

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Unformatted text preview: Chapter 5 - Survey of Probability Concepts Chapter Problem 8 A mortgage holding company has found that 2% of its mortgage mortgage holders default on their mortgage and lose the property. Furthermore, 90% of those who default are late on at least two monthly payments over the life of their mortgage as compared to 45% of those who do not default. What is the joint probability that a mortgagee has two or more late What monthly “defaulting” = 0.02 Define A1 as payments and does not default on the mortgage? Define A2 as "not defaulting" = 1- P(defaulting) = 1 - 0.02 = 0.98 1 2 What is the joint probability that a mortgagee has one or less late What monthly payments and does not default on the mortgage? monthly Define B as “late on at least two payments” and B as “NOT late…” P(BBased P (late on at Theorem, what is the posterior probability that a Based on Bayes' least two payments | default) = 90% 1|A1) = thusortgagee will not least two payments | default) = P(B2 | A1) = 10% the m P (NOT late on at default given one or less late payments over life of the mortgage? P (late on at least two payments | no default) = P( B1 | A2 ) = 0.45 thus P(NOT late on at least two payments | no default) = P (B2 | A2) = 0.55 5-4 Chapter 5 - Survey of Probability Concepts Chapter Event Ai Default Prior Probability P(Ai) .02 Conditional Probability P(B1|Ai) .90 Joint Probability P(Ai and B1) .018 Posterior Probability P(Ai|B1) .018/.459 = .039 No Default .98 .45 .441 (1) .441/.459 = .961 Define A2 as "not defaulting" = 1- P(defaulting) = 1 - 0.02 = 0.98 Define B1 as “at least 2 late payments" P Define P( B1 | A2 ) = .45 (B1) =.459 1.000 P( A2 and B1) = P (A2) P(B1|A2) = 0.98 * 0.45 = 0.441 Chapter 5 - Survey of Probability Concepts Chapter Event Ai Default Prior Probability P(Ai) .02 Conditional Probability P(B1|Ai) .10 Joint Probability P(Ai and B1) .002 Posterior Probability P(Ai|B1) .002/.541 = .004 No Default .98 .55 .539 (2) P (B1) = .541 .539/.541 = .996 (3) 1.000 Define A 2 as "not defaulting" = 1- P(“defaulting”) = 1 - 0.02 = 0.98 Define B as "one or less late payments" P Chapter 5 - Survey of Probability Concepts Chapter Based on Bayes' Theorem, what is the posterior probability that a mortgagee will not default given one or less late payments over the life of the mortgage? “one or less” ≡ “NOT late on at least two…” P (A |B ) = ? Define A1 as “defaulting” = 0.02 Define A2 as "not defaulting" = 1- P(defaulting) = 1 - 0.02 = 0.98 Define B1 as “late on at least two payments” and B2 as “NOT late…” P(B1|A1) = P (late on at least two payments | default) = 90% thus P (NOT late on at least two payments | default) = P(B2 | A1) = 10% P (late on at least two payments | no default) = P( B1 | A2 ) = 0.45 thus P(NOT late on at least two payments | no default) = P (B2 | A2) = 0.55 (.98) * (.55) / [(.02) * (.10) + (.98) * (.55)] = .996 5-4 ...
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