Biostatistics(6)概率与概率分布

3.2.4 相对风险 Relative risk

The relative risk（RR）of B given A is:

RR.png

当A、B两个事件相互独立时，RR=1；当A、B两个事件不相互独立时，则RR不等于1。当非独立性越强，与1的差异则越大。

Example：Relative Risk
Pr(A+)=.1 Pr(B+)=.17 Pr(A+∩B+)=.08
What is the RR of B+ given A+?
Pr(B+|A+)=Pr(B+∩A+)/Pr(A+)=.08/.1=.8
Pr(B+)=Pr(B+∩A+)+Pr(B+∩A-)
Pr(B+∩A-)=Pr(B+)-Pr(B+∩A-)=.17-.08=.09
Pr(B+|A-)=Pr(B+∩A-)/Pr(A-)=Pr(B+∩A-)/.9=.09/.9=.1
RR=Pr(B+|A+)/Pr(B+|A-)=.8/.1=8

3.2.5 贝叶斯公式 Bayes‘ rule

设S为某一试验的样本空间，A为该试验的事件，且P(A)≠0。设B1，B2，...，Bn是S的一个划分，且P(Bi)>0，i=1，2，...,n。则

Bayes' rule.png

特殊地，

Bayes' rule.png

Example of Bayes' rule

The sensitivity of a symptom(or set of symptoms or screening test) is the probability that a symptom is present given that the person has a disease.
敏感性：又称真阳性率，是实际患病且被诊断试验诊为有病的概率
Pr(test+|disease)
The specificity of a symptom(or set of symptoms or screening test) it the probability that the symptom is not present given that the person does not have a disease.
特异性：又称真阴性率，是实际未患病且被诊断试验诊为无病的概率
Pr(test-|no disease)
Let A=symptom and B=disease
Sensitivity=Pr(A|B) Specificity=Pr(A'|B')

The predictive value positive (PV+) of a screening test is the probability that a person has a disease given that the test is positive.
Pr(disease|test+)
阳性预测值:指筛检试验检出的全部阳性例数中，真正“有病”的例数（真阳性）所占的比例，反映筛检试验结果阳性者患目标疾病的可能性
The predictive value negative (PV-) of a screening test is the probability that a person does not have a disease given that the test is negative.
Pr(no disease|test-)
阴性预测值:指检验结果为阴性的受试者中真正未患病的比例。诊断试验的预测值受到敏感度、特异度和受试者中患病率的影响。
Let A=symptom and B=disease
Predictive value positive=PV+=Pr(B|A)
Predictive value negative=PV-=Pr(B'|A')

Let A=symptom and B=disease
PV+=Pr(B|A)=Pr(A|B)Pr(B)/[Pr(A|B)Pr(B)+Pr(A|B')Pr(B')]
即：

PV+.png

where x=Pr(B)=prevalence of disease in the reference population. Similarly,

PV-.png

Hypertension: Suppose 84% of hypertensives and 23%of normotensive are classified as hypertensive by an automated blood-pressure machine. What are the PV+ and PV- of the machine, assuming 20% of the adult population is hypertensive?
Pr(test+|hypertensive)=0.84 即 sensitivity=0.84
Pr(test+|normotensive)=0.23 即 specificity=1-0.23=0.77
Pr(hypertensive)=0.20
PV+=(.84)(.2)/[(.84)(.2)+(.23)(.8)]=.168/.352=.48
PV-=(.77)(.8)/[(.77)(.8)+(.16)(.2)]=.616/.648

Pulmonary Disease:Suppose a 60-year-old man who has never smoked cigarettes presents to a physician with symptoms of a chronic cough and occasional breathlessness. The physician becomes concerned and orders the patient admitted to the hospital for a lung biopsy. Suppose the results fo the lung biopsy are consistent either with lung cancer or with sarcoidosis, a fairly common, nonfatal lung disease. In this case
A={chronic cough,result of lung biopsy}
Disease state:
B1=normal,B2=lung cancer,B3=sarcoidosis
Suppose that Pr(A|B1)=.001,Pr(A|B2)=.9,Pr(A|B3)=.9,
and that in 60-year-old,never smoking men:Pr(B1)=.99,Pr(B2)=.001,Pr(B3)=.009,
then what are the possibilities of the three disease for the 60-year-old man?
Pr(B1|A)=Pr(A|B1)Pr(B1)/[Pr(A|B1)Pr(B1)+Pr(A|B2)Pr(B)+Pr(A|B3)Pr(B3)]=.001(.99)/[.001(.99)+.9(.001)+.9(.009)]=.099
Pr(B2|A)=.9(.001)/[.001(.99)+.9(.001)+.9(.009)]=.090
Pr(B3|A)=.9(.009)/[.001(.99)+.9(.001)+.9(.009)]=.811
Thus, although the unconditional probability of sarcoidosis is very low (.009),the conditional probability of the disease given these symptoms and this age-sex-smoking group is .811. Also, although the symptoms and diagnostic tests are consistent with both lung cancer and sarcoidosis, the latter is much more likely among patients in this age-sex-smoking group.