5.1 Patterns of probable inference  (Page 2/6)

Evidence for a symptom

We let S be the event the symptom is present. The usual case is that the evidence is directly related to the symptom and not the hypothesized condition.The diagnostic test results can say something about an applicant's personal traits, but cannot deal directly with the hypothesized condition. The testresults would be the same whether or not the candidate is successful in the job (he or she does not have the job yet). A geophysical survey deals withcertain structural features beneath the surface. If a fault or a salt dome is present, the geophysical results are the same whether or not there is oilpresent. The physical monitoring report deals with certain physical characteristics. Its reading is the same whether or not the device will fail. A market surveytreats only the condition in the test market. The results depend upon the test market, not the national market. A blood test may be for certainphysical conditions which frequently are related (at least statistically) to the disease. But the result of the blood test for the physical conditionis not directly affected by the presence or absence of the disease.

Under conditions of this type, we may assume

These imply $\{E,H\}\mathrm{ci}|S$ and $\mathrm{ci}|S^c$ . Now

It is worth noting that each term in the denominator differs from the corresponding term in the numerator by having H ^c in place of H . Before completing the analysis, it is necessary to consider how H and S are related stochastically in the data. Four cases may be considered.

1. Data are $P\left(S\right|H)$ , $P\left(S\right|H^c)$ , and $P\left(H\right)$ .
2. Data are $P\left(S\right|H)$ , $P\left(S\right|H^c)$ , and $P\left(S\right)$ .
3. Data are $P\left(H\right|S)$ , $P\left(H\right|S^c)$ , and $P\left(S\right)$ .
4. Data are $P\left(H\right|S)$ , $P\left(H\right|S^c)$ , and $P\left(H\right)$ .

• Case a
  $$\frac{P\left(H\right|E)}{P\left(H^c\right|E)}=\frac{P\left(H\right)P\left(S\right|H)P\left(E\right|S)+P\left(H\right)P\left(S^c\right|H)P\left(E\right|S^c)}{P\left(H^c\right)P\left(S\right|H^c)P\left(E\right|S)+P\left(H^c\right)P\left(S^c\right|H^c)P\left(E\right|S^c)}$$

  Geophysical survey

  Let H be the event of a successful oil well, S be the event there is a geophysical structure favorable to the presence of oil, and E be the event the geophysical survey indicates a favorable structure. We suppose $\{H,E\}\mathrm{ci}|S$ and $\mathrm{ci}|S^c$ . Data are

  $$P\left(H\right)/P\left(H^c\right)=3,P\left(S\right|H)=0.92,P\left(S\right|H^c)=0.20,P\left(E\right|S)=0.95,P\left(E\right|S^c)=0.15$$

  Then

  $$\frac{P\left(H\right|E)}{P\left(H^c\right|E)}=3·\frac{0.92·0.95+0.08·0.15}{0.20·0.95+0.80·0.15}=\frac{1329}{155}=8.5742$$

  $$\text{so}\text{that}P\left(H\right|E)=1-\frac{155}{1484}=0.8956$$

  The geophysical result moved the prior odds of 3/1 to posterior odds of 8.6/1, with acorresponding change of probabilities from 0.75 to 0.90.

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• Case b Data are $P\left(S\right)$ $P\left(S\right|H)$ , $P\left(S\right|H^c)$ , $P\left(E\right|S)$ . and $P\left(E\right|S^c)$ . If we can determine $P\left(H\right)$ , we can proceed as in case a. Now by the law of total probability
  $$P\left(S\right)=P\left(S\right|H)P\left(H\right)+P\left(S\right|H^c)[1-P\left(H\right)]$$
  which may be solved algebraically to give
  $$P\left(H\right)=\frac{P\left(S\right)-P\left(S\right|H^c)}{P\left(S\right|H)-P\left(S\right|H^c)}$$

  Geophysical survey revisited

  In many cases a better estimate of $P\left(S\right)$ or the odds $P\left(S\right)/P\left(S^c\right)$ can be made on the basis of previous geophysical data. Suppose the prior odds for S are 3/1, so that $P\left(S\right)=0.75$ . Using the other data in [link] , we have

  $$P\left(H\right)=\frac{P\left(S\right)-P\left(S\right|H^c)}{P\left(S\right|H)-P\left(S\right|H^c)}=\frac{0.75-0.20}{0.92-0.20}=55/72,\text{so}\text{that}\frac{P\left(H\right)}{P\left(H^c\right)}=55/17$$

  Using the pattern of case a, we have

  $$\frac{P\left(H\right|E)}{P\left(H^c\right|E)}=\frac{55}{17}·\frac{0.92·0.95+0.08·0.15}{0.20·0.95+0.80·0.15}=\frac{4873}{527}=9.2467$$

  $$\text{so}\text{that}P\left(H\right|E)=1-\frac{527}{5400}=0.9024$$

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  Usually data relating test results to symptom are of the form $P\left(E\right|S)$ and $P\left(E\right|S^c)$ , or equivalent. Data relating the symptom and the hypothesized condition may go either way. Incases a and b, the data are in the form $P\left(S\right|H)$ and $P\left(S\right|H^c)$ , or equivalent, derived from data showing the fraction of times the symptom is noted when the hypothesizedcondition is identified. But these data may go in the opposite direction, yielding $P\left(H\right|S)$ and $P\left(H\right|S^c)$ , or equivalent. This is the situation in cases c and d.
• Case c Data are $P\left(E\right|S),P\left(E\right|S^c),P\left(H\right|S),P\left(H\right|S^c)$ and $P\left(S\right)$ .

  Evidence for a disease symptom with prior $P\left(S\right)$

  When a certain blood syndrome is observed, a given disease is indicated 93 percent of the time. The disease is found without this syndrome only three percent of the time. Atest for the syndrome has probability 0.03 of a false positive and 0.05 of a false negative. A preliminary examination indicates a probability 0.30 that a patienthas the syndrome. A test is performed; the result is negative. What is the probability the patient has the disease?

  Solution

  In terms of the notation above, the data are

  $$P\left(S\right)=0.30,P\left(E\right|S^c)=0.03,P\left(E^c\right|S)=0.05,$$

  $$P\left(H\right|S)=0.93,\text{and}P\left(H\right|S^c)=0.03$$

  We suppose $\{H,E\}\mathrm{ci}|S$ and $\mathrm{ci}|S^c$ .

  $$\frac{P\left(H\right|E^c)}{P\left(H^c\right|E^c)}=\frac{P\left(S\right)P\left(H\right|S)P\left(E^c\right|S)+P\left(S^c\right)P\left(H\right|S^c)P\left(E^c\right|S^c)}{P\left(S\right)P\left(H^c\right|S)P\left(E^c\right|S)+P\left(S^c\right)P\left(H^c\right|S^c)P\left(E^c\right|S^c)}$$

  $$=\frac{0.30·0.93·0.05+0.70·0.03·0.97}{0.30·0.07·0.05+0.70·0.97·0.97}=\frac{429}{8246}$$

  which implies $P\left(H\right|E^c)=429/8675\approx 0.05$ .

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• Case d This differs from case c only in the fact that a prior probability for H is assumed. In this case, we determine the corresponding probability for S by
  $$P\left(S\right)=\frac{P\left(H\right)-P\left(H\right|S^c)}{P\left(H\right|S)-P\left(H\right|S^c)}$$
  and use the pattern of case c.

  Evidence for a disease symptom with prior $P\left(H\right)$

  Suppose for the patient in [link] the physician estimates the odds favoring the presence of the disease are 1/3, so that $P\left(H\right)=0.25$ . Again, the test result is negative. Determine the posterior odds, given E ^c .

  Solution

  First we determine

  $$P\left(S\right)=\frac{P\left(H\right)-P\left(H\right|S^c)}{P\left(H\right|S)-P\left(H\right|S^c)}=\frac{0.25-0.03}{0.93-0.03}=11/45$$

  Then

  $$\frac{P\left(H\right|E^c)}{P\left(H^c\right|E^c)}=\frac{(11/45)·0.93·0.05+(34/45)·0.03·0.97}{(11/45)·0.07·0.05+(34/45)·0.97·0.97}=\frac{15009}{320291}=0.047$$

  The result of the test drops the prior odds of 1/3 to approximately 1/21.

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