Bayesian Theorem

11 Feb 2018 on Math

1. state of theorem

$P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}}$

- prior probability

$P(A)$

the initial degree of belief in A

- likelihood probability

$P(B\mid A)$

- poterior probability

$P(A\mid B)$

the probability for A after taking into account B for and against A

2. conditaional probability

$P(A\mid B)$

conditional probability, the likelihood of event A occurring given that B is true

$P(B\mid A)$

conditional probability, the likelihood of event B occurring given that A is true

$P(A) and {\displaystyle P(B)}$

marginal probability

memorable form

$P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}}}=\frac{(likelihood * prior)$

$P(A\mid B)\propto P(A)\cdot P(B\mid A)$

3. drug test example

Suppose that a test for using a particular drug is 99% sensitive and 99% specific. That is, the test will produce 99% true positive results for drug users and 99% true negative results for non-drug users. Suppose that 0.5% of people are users of the drug. What is the probability that a randomly selected individual with a positive test is a drug user?

${\begin{aligned}P({\text{User}}\mid {\text{+}})&={\frac {P({\text{+}}\mid {\text{User}})P({\text{User}})}{P(+)}}\\&={\frac {P({\text{+}}\mid {\text{User}})P({\text{User}})}{P({\text{+}}\mid {\text{User}})P({\text{User}})+P({\text{+}}\mid {\text{Non-user}})P({\text{Non-user}})}}\\[8pt]&={\frac {0.99\times 0.005}{0.99\times 0.005+0.01\times 0.995}}\\[8pt]&\approx 33.2\%\end{aligned}}$