Updating Disease Risk After Independent Tests

Why Combine Test Results?

A diagnostic test changes how plausible disease is, but a second test can change that assessment again. The calculation depends on the starting disease probability and on how likely the observed results are among people with and without the disease. When the tests are conditionally independent given disease status, their likelihoods can be multiplied within each of those two groups.

The word conditionally matters. The independence condition is applied separately among people who have the disease and among people who do not. It does not generally say that the test results are independent in the full population. In the examples below, all data are hypothetical, and the conditional-independence assumption is stated explicitly before it is used.

Model and Updating Formula

Definition: Let $D$ denote the event that a person has a particular disease, and let $D^c$ denote the event that the person does not. Two test results $E_1$ and $E_2$ are conditionally independent given disease status when $$ P(E_1\cap E_2\mid D)=P(E_1\mid D)P(E_2\mid D) $$ and $$ P(E_1\cap E_2\mid D^c)=P(E_1\mid D^c)P(E_2\mid D^c). $$ Here, $E_1$ and $E_2$ are the particular results observed, such as a positive result on the first test and a negative result on the second.

Write the prior disease probability as $\pi=P(D)$. For the observed pair of results, define the diseased-case likelihood $L_D=P(E_1\cap E_2\mid D)$ and the non-diseased-case likelihood $L_{D^c}=P(E_1\cap E_2\mid D^c)$. Under conditional independence, each is a product:

L_D=P(E_1\mid D)P(E_2\mid D),\qquad L_{D^c}=P(E_1\mid D^c)P(E_2\mid D^c).

Bayes’ rule then gives the updated probability of disease after both results:

P(D\mid E_1\cap E_2) = \frac{\pi L_D}{\pi L_D+(1-\pi)L_{D^c}}.

The denominator is the total probability of observing that pair of results: the probability that a person has the disease and produces the results, plus the probability that a person does not have the disease and produces them. This is why the calculation needs likelihoods for both disease-status groups, not just the chance of the results among diseased people.

Where Sensitivity and Specificity Enter

For a test with positive and negative outcomes, sensitivity is the probability of a positive result among people with disease, $P(+\mid D)$. Specificity is the probability of a negative result among people without disease, $P(-\mid D^c)$. Thus, a negative result among diseased people has probability $1-\text{sensitivity}$, while a positive result among non-diseased people has probability $1-\text{specificity}$. For two tests, use the probability matching each observed outcome and disease-status group.

For example, if the observed results are positive on test 1 and negative on test 2, the diseased-case likelihood is the sensitivity of test 1 multiplied by one minus the sensitivity of test 2. The non-diseased-case likelihood is one minus the specificity of test 1 multiplied by the specificity of test 2. This pairing of outcomes with the correct conditional probabilities is essential.

Worked Example: Two Positive Results When Disease Is Uncommon

Suppose a hypothetical disease has prevalence $\pi=0.02$ in a population. Test 1 has sensitivity $0.90$ and specificity $0.95$; test 2 has sensitivity $0.80$ and specificity $0.90$. A person tests positive on both. Assume the results are conditionally independent given disease status. Find the probability that the person has the disease after these results.

For someone with the disease, the probability of both positive results is $$ L_D=P(+_1\mid D)P(+_2\mid D)=0.90(0.80)=0.72. $$

For someone without the disease, the false-positive probabilities are $1-0.95=0.05$ and $1-0.90=0.10$. Therefore, $$ L_{D^c}=P(+_1\mid D^c)P(+_2\mid D^c)=0.05(0.10)=0.005. $$

Applying Bayes’ rule, $$ P(D\mid +_1,+_2) =\frac{0.02(0.72)}{0.02(0.72)+0.98(0.005)} =\frac{0.0144}{0.0144+0.0049} =\frac{0.0144}{0.0193} \approx 0.7461. $$

Under the stated model, the probability of disease after both positive results is about $74.6\%$. Although the two results make disease much more plausible than the $2\%$ prior probability, the posterior is not $100\%$: people without disease can also produce two positive results.

Why Multiplying Likelihoods Works

The multiplication step follows directly from conditional independence. The posterior probability is the joint probability of disease and the observed results divided by the probability of those results. The numerator can be written as the prior probability of disease times the probability of the results among diseased people. The denominator splits into the corresponding contributions from diseased and non-diseased people.

Write Bayes’ rule for the pair of results.
For $E=E_1\cap E_2$, $P(D\mid E)=P(D)P(E\mid D)/P(E)$.

Expand the denominator by disease status.
The law of total probability gives $P(E)=P(D)P(E\mid D)+P(D^c)P(E\mid D^c)$.

Use conditional independence.
Replace $P(E\mid D)$ by $P(E_1\mid D)P(E_2\mid D)$, and replace $P(E\mid D^c)$ by $P(E_1\mid D^c)P(E_2\mid D^c)$.

Substitute the prior probabilities.
Since $P(D)=\pi$ and $P(D^c)=1-\pi$, the resulting expression is $\pi L_D/[\pi L_D+(1-\pi)L_{D^c}]$.

The same calculation has an odds form. When all relevant probabilities are positive, define the likelihood ratio for a result $E_i$ as $P(E_i\mid D)/P(E_i\mid D^c)$. Dividing the posterior probability by its complement gives

\frac{P(D\mid E_1,E_2)}{P(D^c\mid E_1,E_2)} = \frac{\pi}{1-\pi} \left(\frac{P(E_1\mid D)}{P(E_1\mid D^c)}\right) \left(\frac{P(E_2\mid D)}{P(E_2\mid D^c)}\right).

Thus, under the conditional-independence model, posterior odds equal prior odds multiplied by the two result-specific likelihood ratios. A result is evidence in favor of disease when its likelihood is larger among diseased people than among non-diseased people; the size and direction of the update depend on that ratio.

Worked Example: A Positive Result Followed by a Negative Result

Suppose a hypothetical disease has prevalence $0.10$. Test 1 has sensitivity $0.95$ and specificity $0.90$; test 2 has sensitivity $0.80$ and specificity $0.95$. A person tests positive on test 1 and negative on test 2. Assume conditional independence given disease status. Calculate the updated disease probability.

Among people with disease, the probability of this mixed result is the probability of a positive result on test 1 times the probability of a negative result on test 2: $$ L_D=0.95(1-0.80)=0.95(0.20)=0.19. $$

Among people without disease, a positive result on test 1 has probability $1-0.90=0.10$, and a negative result on test 2 has probability $0.95$. Therefore, $$ L_{D^c}=0.10(0.95)=0.095. $$

The posterior probability is $$ P(D\mid +_1,-_2) =\frac{0.10(0.19)}{0.10(0.19)+0.90(0.095)} =\frac{0.019}{0.019+0.0855} =\frac{0.019}{0.1045} \approx 0.1818. $$

The resulting disease probability is about $18.2\%$, compared with the prior $10\%$. In this example, the positive first result raises the probability, and the negative second result reduces some of that evidence. The final probability reflects both results rather than treating either one alone as decisive.

Negative Results and Residual Risk

The same formula applies when both results are negative. The diseased-case likelihood then uses the false-negative probability for each test, while the non-diseased-case likelihood uses the probability of a negative result for each test. A small false-negative probability can make a negative result strong evidence against disease, but the final probability also depends on the prior prevalence and on the other test’s characteristics.

Worked Example: Two Negative Results

Suppose a hypothetical disease has prevalence $0.15$. Test 1 has sensitivity $0.90$ and specificity $0.95$; test 2 has sensitivity $0.85$ and specificity $0.90$. A person tests negative on both. Assume the results are conditionally independent given disease status. Find the residual disease probability.

For a person with disease, the false-negative probabilities are $1-0.90=0.10$ and $1-0.85=0.15$, so $$ L_D=0.10(0.15)=0.015. $$

For a person without disease, the probabilities of negative results are the specificities: $$ L_{D^c}=0.95(0.90)=0.855. $$

Therefore, $$ P(D\mid -_1,-_2) =\frac{0.15(0.015)}{0.15(0.015)+0.85(0.855)} =\frac{0.00225}{0.00225+0.72675} =\frac{0.00225}{0.729} \approx 0.003086. $$

The residual probability of disease is about $0.309\%$ under these assumptions. It is lower than the prior $15\%$, but it is not exactly zero. A negative result reduces risk; it does not logically rule out disease.

Assumptions and Common Pitfalls

The multiplication of the two likelihoods is justified only if conditional independence holds for the observed outcomes within both disease-status groups. The formula also requires that the prior probability and conditional test probabilities apply to the person or population being assessed. In practice, these quantities come from an appropriate source and may not be known exactly. The arithmetic can be correct while the assumptions or input probabilities are unsuitable.

Multiplying probabilities from the wrong group. For a positive result among people without disease, use one minus specificity, not specificity. For a negative result among people with disease, use one minus sensitivity.
Multiplying marginal result probabilities. Conditional independence supports multiplication within the diseased group and within the non-diseased group. The posterior calculation then combines those two likelihoods with the prior probabilities.
Ignoring the prior. Strong test performance alone does not determine the posterior probability. The prevalence, or relevant prior disease probability, affects how many non-diseased people contribute possible test-result patterns.
Calling the posterior a test accuracy. A posterior such as $P(D\mid +_1,+_2)$ is the probability of disease given those observed results under the model. It is not the sensitivity, which is $P(+\mid D)$ for a single test.

A common incorrect argument is: “Both tests were positive, so the disease probability is the product of their sensitivities.” That product is only the probability of two positive results among people with disease under conditional independence. It omits the prior probability and the probability of the same results among people without disease. The correction is to compute both likelihoods and substitute them into Bayes’ rule.

Key takeaway: For two conditionally independent test results, multiply their conditional probabilities separately among diseased and non-diseased people, then combine the two likelihoods with the prior disease probability using Bayes’ rule.

Check Your Understanding

For each question, distinguish the conditional likelihoods from the final posterior probability.

In a two-test model, what two conditional-independence statements are needed to multiply the likelihoods for a particular pair of results?
A test has sensitivity $0.88$ and specificity $0.94$. What is the probability of a negative result among people with disease, and what is the probability of a positive result among people without disease?
Why does calculating $P(E_1\mid D)P(E_2\mid D)$ alone not give the probability of disease after observing both results?
If both tests are negative, which conditional probabilities are multiplied to form the likelihood among people with disease? Which are multiplied among people without disease?
Explain in words what $P(D\mid E_1\cap E_2)$ means, and why it is not the same quantity as test sensitivity.