Multiple-Test Models With Conditional Independence

Why Multiple Test Results Need Careful Conditioning

A person who takes several diagnostic tests may receive a pattern such as positive, negative, positive. To evaluate that pattern, it is not enough to multiply the tests’ separate positive or negative probabilities automatically. A common model assumes that test results are independent given the person’s true disease status. The results may nevertheless be dependent when disease status is unknown, because they share that status as a common source of information.

This distinction matters whenever several measurements are used together. Conditional independence lets us calculate a pattern probability within each disease-status group; the law of total probability then combines those groups, and Bayes’ theorem updates the probability of disease after observing the pattern. The calculations below use hypothetical diagnostic tests and populations.

Definition: Let \(D\) denote the event that a person has a disease, and let \(T_j\) denote the result of test \(j\). The tests are conditionally independent given disease status if, for either \(H=D\) or \(H=D^c\), the probability of any specified combination of results is the product of the corresponding probabilities conditional on \(H\). For two tests, for example, \(P(T_1=+,T_2=+\mid H)=P(T_1=+\mid H)P(T_2=+\mid H)\). This assumption does not assert that the tests are independent without conditioning.

The Multiple-Test Calculation

Suppose \(n\) tests are conditionally independent given disease status. Write \(\pi=P(D)\) for disease prevalence, \(s_j=P(T_j=+\mid D)\) for the sensitivity of test \(j\), and \(f_j=P(T_j=+\mid D^c)\) for its false-positive probability. Thus, the specificity of test \(j\) is \(1-f_j\). For a particular observed pattern, define \(r_j\) as the conditional probability of the result on test \(j\), with the disease status specified. A positive result contributes \(s_j\) in the disease group and \(f_j\) in the no-disease group; a negative result contributes \(1-s_j\) and \(1-f_j\), respectively.

The product of these contributions gives the probability of the pattern within each group. The unconditional probability is a prevalence-weighted combination of those two conditional probabilities. Bayes’ theorem then gives the probability of disease after observing the pattern.

P(\text{pattern})=\pi\prod_{j=1}^{n}P(T_j=\text{observed result}_j\mid D) +(1-\pi)\prod_{j=1}^{n}P(T_j=\text{observed result}_j\mid D^c)

P(D\mid \text{pattern})= \frac{\pi\prod_{j=1}^{n}P(T_j=\text{observed result}_j\mid D)} {\pi\prod_{j=1}^{n}P(T_j=\text{observed result}_j\mid D) +(1-\pi)\prod_{j=1}^{n}P(T_j=\text{observed result}_j\mid D^c)}

These formulas require that the tests be conditionally independent within both status groups, and that the stated probabilities apply to the population being modeled. If, for example, test errors remain associated within the disease group because the tests share a measurement process, the conditional products need not be valid. In that case, the joint conditional probabilities must be modeled directly.

Worked Example: Two Positive Results in a Low-Prevalence Population

In a hypothetical screening program, disease prevalence is \(0.10\). Two tests each have sensitivity \(0.80\) and specificity \(0.90\). Assume the tests are conditionally independent given disease status. Find (a) the probability that both results are positive and (b) the probability of disease given that both are positive.

Method and assumptions. The false-positive probability for each test is \(1-0.90=0.10\). Conditional independence allows multiplication within the disease and no-disease groups. The prior probabilities are \(P(D)=0.10\) and \(P(D^c)=0.90\).

Both positive if disease is present: \(P(T_1=+,T_2=+\mid D)=0.80(0.80)=0.64\).

Both positive if disease is absent: \(P(T_1=+,T_2=+\mid D^c)=0.10(0.10)=0.01\).

(a) Unconditional probability: Combine the two cases using total probability: \(P(T_1=+,T_2=+)=0.10(0.64)+0.90(0.01)=0.064+0.009=0.073\). To four decimal places, this is \(0.0730\).

(b) Posterior probability: The joint probability of disease and two positive results is \(0.10(0.64)=0.064\). Therefore, \(P(D\mid T_1=+,T_2=+)=0.064/0.073=0.876712\ldots\), or \(0.8767\) to four decimal places.

Check and common mistake. Each test’s marginal positive probability is \(0.10(0.80)+0.90(0.10)=0.17\). Multiplying these marginals gives the incorrect result \(0.17^2=0.0289\) if used as the probability of two positives. The correct probability is \(0.0730\), found by conditioning on status first. The two calculations differ because conditional independence does not imply unconditional independence.

Why Conditional Independence Can Disappear After Mixing

For two tests, the distinction can be shown directly. Let \(A\) be the event that test 1 is positive and \(B\) the event that test 2 is positive. Suppose conditional independence holds given disease status. Write \(a_1=P(A\mid D)\), \(a_2=P(B\mid D)\), \(b_1=P(A\mid D^c)\), and \(b_2=P(B\mid D^c)\). With prevalence \(\pi=P(D)\), total probability gives the joint probability and both marginal probabilities as follows.

P(A\cap B)=\pi a_1a_2+(1-\pi)b_1b_2,\qquad P(A)=\pi a_1+(1-\pi)b_1,\qquad P(B)=\pi a_2+(1-\pi)b_2

Subtracting the product of the marginals from the joint probability gives the unconditional covariance of the two event indicators. Expanding the product makes the cancellation explicit:

\begin{aligned} P(A\cap B)-P(A)P(B) &=\pi a_1a_2+(1-\pi)b_1b_2\\ &\quad-[\pi a_1+(1-\pi)b_1][\pi a_2+(1-\pi)b_2]\\ &=\pi(1-\pi)(a_1-b_1)(a_2-b_2). \end{aligned}

For tests that are more likely to be positive when disease is present, \(a_1>b_1\) and \(a_2>b_2\). If \(0<\pi<1\), the expression is positive, so the positive-result events are positively associated overall even though they are independent within each disease-status group. Observing one positive result makes disease more plausible, which in turn makes a second positive result more plausible.

Specify the observed pattern.
Record whether each test is positive or negative; a negative result uses the complement of the relevant positive probability.

Calculate its conditional probability in each group.
Multiply the test-result probabilities within \(D\), and separately within \(D^c\), using conditional independence.

Combine the groups.
Weight the two conditional pattern probabilities by \(\pi\) and \(1-\pi\) to obtain the unconditional pattern probability.

Update if disease probability is requested.
Divide the disease-group contribution by the unconditional pattern probability.

Worked Example: Updating With a Mixed Three-Test Pattern

A hypothetical population has disease prevalence \(0.05\). Three tests are conditionally independent given disease status. Their sensitivities are \(0.90\), \(0.80\), and \(0.70\); their false-positive probabilities are \(0.10\), \(0.20\), and \(0.05\). Find the probability of the pattern positive, negative, positive, and then find the probability of disease given that pattern.

Method. Use a product for the three results within each status group, then apply total probability and Bayes’ theorem. The prior probabilities are \(P(D)=0.05\) and \(P(D^c)=0.95\).

Pattern probability conditional on disease: The positive, negative, positive pattern has probability \(0.90(1-0.80)(0.70)=0.90(0.20)(0.70)=0.126\).

Pattern probability conditional on no disease: It has probability \(0.10(1-0.20)(0.05)=0.10(0.80)(0.05)=0.004\).

Unconditional pattern probability: \(0.05(0.126)+0.95(0.004)=0.0063+0.0038=0.0101\).

Posterior probability: The disease-and-pattern contribution is \(0.05(0.126)=0.0063\), so \(P(D\mid +,-,+)=0.0063/0.0101=0.623762\ldots\). To four decimal places, the posterior is \(0.6238\).

Check and common mistake. The two status-specific contributions sum to \(0.0063+0.0038=0.0101\), the probability of the observed pattern. Using the three marginal probabilities as though they were unconditionally independent would ignore the shared disease status and would not implement the stated model.

When the Observation Is a Summary of Results

Sometimes the information is not the exact order of results, but a description such as “exactly one test was positive.” That description corresponds to several disjoint patterns. Calculate each pattern’s conditional probability and add them within each status group before applying total probability and Bayes’ theorem. Do not treat the summary event as if it were one particular ordered pattern.

Worked Example: Exactly One Positive Among Two Tests

In a hypothetical population, disease prevalence is \(0.10\). Two tests are conditionally independent given disease status. Test 1 has sensitivity \(0.90\) and false-positive probability \(0.10\); test 2 has sensitivity \(0.80\) and false-positive probability \(0.05\). Find the probability of exactly one positive result and the probability of disease given exactly one positive.

Method. Exactly one positive consists of the disjoint patterns \(+,-\) and \(-,+\). Add their probabilities separately within the disease and no-disease groups.

Conditional on disease: \(P(+,-\mid D)=0.90(1-0.80)=0.90(0.20)=0.18\), and \(P(-,+\mid D)=(1-0.90)(0.80)=0.10(0.80)=0.08\). Thus \(P(\text{exactly one }+\mid D)=0.18+0.08=0.26\).

Conditional on no disease: \(P(+,-\mid D^c)=0.10(1-0.05)=0.10(0.95)=0.095\), and \(P(-,+\mid D^c)=(1-0.10)(0.05)=0.90(0.05)=0.045\). Thus \(P(\text{exactly one }+\mid D^c)=0.095+0.045=0.14\).

Unconditional probability: \(0.10(0.26)+0.90(0.14)=0.026+0.126=0.152\).

Posterior probability: \(P(D\mid\text{exactly one }+)=0.026/0.152=0.171052\ldots\), or \(0.1711\) to four decimal places. As a check, \(0.026+0.126=0.152\), matching the unconditional probability.

Common mistake avoided. Using only the \(+,-\) pattern would omit the disjoint \(-,+\) case. The event “exactly one positive” includes both, so both contributions must be included in the numerator and denominator.

Assumptions and Exam Traps

The calculations are reliable only when the model matches the situation. “Conditionally independent” is a substantive assumption about test behavior within each disease-status group; it is not guaranteed merely because different tests were used. A shared sample, a common instrument, or a common source of measurement error can create association even after conditioning on disease status. If such dependence is specified, use the joint conditional probabilities rather than multiplying individual probabilities.

A useful check is to keep the status groups separate until the last step. For a posterior probability, the numerator must be the probability of both disease and the observed evidence. The denominator must include the evidence under both disease and no disease. The following errors are particularly common:

Multiplying marginal probabilities. This assumes unconditional independence, which does not follow from conditional independence.
Using specificity as the false-positive probability. If specificity is \(c\), the probability of a positive result without disease is \(1-c\).
Using a positive-result probability for a negative result. Conditional on disease, a negative result has probability \(1-s_j\), not \(s_j\); conditional on no disease, it has probability \(1-f_j\).
Forgetting a pattern component. Every test result in the observed pattern contributes a factor, and summary events such as “exactly one positive” require adding all included patterns.
Updating with only the likelihood. A disease posterior includes the prevalence-weighted disease contribution and the corresponding no-disease contribution in its denominator.

Key takeaway: Under conditional independence, multiply test-result probabilities only after fixing disease status. Weight the resulting pattern probabilities by prevalence to find an unconditional probability, and use the disease-group contribution divided by the total to obtain the posterior probability.

Check Your Understanding

For each problem, assume conditional independence given disease status where stated. Give probabilities to four decimal places.

A hypothetical screening group has prevalence \(0.04\). Two tests each have sensitivity \(0.85\) and false-positive probability \(0.10\), and are conditionally independent given status. Find the probability both are positive and the posterior probability of disease given two positives. (Answer: Joint probability \(0.0385\); posterior \(0.7506\).)
A hypothetical population has prevalence \(0.20\). Test A has sensitivity \(0.90\) and false-positive probability \(0.05\); test B has sensitivity \(0.70\) and false-positive probability \(0.10\). Given conditional independence, find the probability of the pattern \(+,-\) and the posterior probability of disease. (Answer: Pattern probability \(0.0900\); posterior \(0.6000\).)
A hypothetical population has prevalence \(0.03\). Three conditionally independent tests have sensitivities \(0.80, 0.90, 0.75\) and false-positive probabilities \(0.10, 0.05, 0.20\), respectively. Find the probability and disease posterior for the pattern \(-,+,-\). (Answer: Pattern probability \(0.0363\); posterior \(0.0372\).)
A hypothetical population has prevalence \(0.25\). Two conditionally independent tests have sensitivities \(0.80\) and \(0.60\), and false-positive probabilities \(0.20\) and \(0.10\). Find the joint probability that both tests are positive, each marginal positive probability, and the product of the marginals. (Answer: Joint probability \(0.1350\); marginals \(0.3500\) and \(0.2250\); product \(0.0788\), so the positive-result events are not unconditionally independent.)
A hypothetical population has prevalence \(0.10\). Two conditionally independent tests have sensitivities \(0.90\) and \(0.80\), and false-positive probabilities \(0.10\) and \(0.05\). Find the probability of exactly one positive result and the posterior probability of disease given exactly one positive. (Answer: Probability \(0.1520\); posterior \(0.1711\).)

What You'll Learn