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  • STA256 Lecture 04

Bayes' Theorem

• We have $\{B_1,B_2,\dots ,B_k\}$ as partition of $S$
• $P(B_j|A)=\frac{P(A|B_j)P(B_j)}{\sum _{i=1}^{k}P(A|B_i)P(B_i)}$
• From the Law of Total Probability.
• $P(A|B_j)=\frac{P(A\cap B_j)}{P(B_j)}$
  • So $P(A\cap B_j)=P(A|B_j)\cdot P(B_j)$
• So it follows that $P(B_j|A)=\frac{P(B_j\cap A)}{P(A)}$
  • $P(B_j|A)\cdot P(A)=P(B_j\cap A)$
  • $P(A|B_j)\cdot P(B_j)=P(A\cap B_j)$
  • So with the two above:
    • $P(A|B_j)\cdot P(B_j)=P(B_j|A)\cdot P(A)$
    • $\frac{P(A|B_j)\cdot P(B_j)}{P(A)}=P(B_j|A)$
    • From theorem 20, we know $P(A)=\sum P(B_j\cdot P(A|B_j))$
    • $P(B_j|A)=\frac{P(A|B_j)\cdot P(B_j)}{\sum P(B_j)\cdot P(A|B_j)}$
• Example:
  • Slide 51
  • $U_{+}=\text{Uses Drugs}$
  • $U_{-}=\text{Doesn’t Use Drugs}$
  • $T_{+}=\text{Positve Test Results}$
  • $T_{-}=\text{Negative Test Results}$
  • Randomly chosen, what's the chance they're positive?
    • $P(T_{+})=P(U_{+})\cdot P(T_{+}|U_{+})+P(U_{-})\cdot P(T_{+}|U_{-})$
    • $(0.01)(0.99)+(0.99)(0.005)$
    • $0.01485$
    • $1.485\mathrm{\%}$
  • What's the chance that a positive drug test correctly signals a drug user: Application of Bayes' Theorem
    • $P(U_{+}|T_{+})=\frac{P(U_{+}\cap T_{+})}{P(T_{+})}$
    • $P(U_{+}|T_{+})=\frac{P(0.01\cdot 0.99)}{P(U_{+})\cdot P(T_{+}|U_{+})+P(U_{-})\cdot P(T_{+}|U_{-})}$
    • $P(U_{+}|T_{+})=\frac{0.01\cdot 0.99}{0.01485}$
    • $P(U_{+}|T_{+})=\frac{0.0099}{0.01485}$
    • $P(U_{+}|T_{+})=0.6666666667$
    • $P(U_{+}|T_{+})=\frac{2}{3}$
    • So only $\frac{2}{3}$rds of the time it's correct.