Independence

In probability, two events are said to be independent if the probability of one is not affected by the occurrence or non-occurrence of the other. This definition requires further explanation, so consider the following example.

Earlier in this module we considered data from a population of N=100 men who had both a PSA test and a biopsy for prostate cancer. Suppose we have a different test for prostate cancer. This prostate test produces a numerical risk that classifies a man as at low, moderate or high risk for prostate cancer. A sample of 100 men underwent the new test and also had a biopsy. The data from the biopsy results are summarized below.

Prostate Test Risk

Prostate Cancer

No Prostate Cancer

Total

Low

10

50

60

Moderate

6

30

36

High

4

20

24

 

20

100

120

Note that regardless of whether the hypothetical Prostate Test was low, moderate, or high, the probability that a subject had cancer was 0.167. In other words, knowing a man's prostate test result does not affect the likelihood that he has prostate cancer in this example. In this case, the probability that a man has prostate cancer is independent of his prostate test result.

Demonstrating Independence

Consider two events, call them A and B (e.g., A might be a low risk based on the "prostate test", and B is a diagnosis of prostate cancer). These two events are independent if P(A | B) = P(A) or if P(B | A) = P(B).

To check independence, we compare a conditional and an unconditional probability: P(A | B) = P(Low Risk | Prostate Cancer) = 10/20 = 0.50 and P(A) = P(Low Risk) = 60/120 = 0.50. The equality of the conditional and unconditional probabilities indicates independence.

Independence can also be tested by examining whether P(B | A) = P(Prostate Cancer | Low Risk) = 10/60 = 0.167 and P(B) = P(Prostate Cancer) = 20/120 = 0.167. In other words, the probability of the patient having a diagnosis of prostate cancer given a low risk "prostate test" (the conditional probability) is the same as the overall probability of having a diagnosis of prostate cancer (the unconditional probability).

Example:

The following table contains information on a population of N=6,732 individuals who are classified as having or not having prevalent cardiovascular disease (CVD). Each individual is also classified in terms of having a family history of cardiovascular disease. In this analysis, family history is defined as a first degree relative (parent or sibling) with diagnosed cardiovascular disease before age 60.

 

Prevalent CVD

Free of CVD

Total

Family History of CVD

491

368

859

No Family History of CVD

152

5,721

5,873

Total

643

6,089

6,732

Are family history and prevalent CVD independent? Is there a relationship between family history and prevalent CVD? This is a question of independence of events.

Let A=Prevalent CVD and B = Family History of CVD. (Note that it does not matter how we define A and B, for example we could have defined A=No Family History and B=Free of CVD, the result will be identical.) We now must check whether P(A | B) = P(A) or if P(B | A) = P(B). Again, it makes no difference which definition is used; the conclusion will be identical. We will compare the conditional probability to the unconditional probability as follows:

Conditional Probability

Unconditional Probability

P(A | B) = P(Prevalent CVD | Family History of CVD) = 491/859 = 0.572

 

The probability of prevalent CVD given a family history is 57.2% (as compared to 2.6% among patients with no family history).

P(A) = P(Prevalent CVD) = 643/6,732 = 0.096

 

In the overall population, the probability of prevalent CVD is 9.6% (or 9.6% of the population has prevalent CVD).

Since these probabilities are not equal, family history and prevalent CVD are not independent. Individuals with a family history of CVD are much more likely to have prevalent CVD. 


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