Why Repeated Tests May Not Be Independent

Why a Repeat Result May Add Less Evidence

Suppose a screening test is positive, and a repeat of the same test is positive too. It is tempting to treat the two positives as two separate pieces of evidence and multiply the likelihood of a positive result by itself. That calculation is justified only under a particular condition: the two results must be independent within each disease-status group. If both tests are influenced by a shared biological or technical factor, the second result may be less surprising after the first than an independence calculation assumes.

For example, two tests on the same specimen can share an unusually high concentration of a target, a contaminant, or an interfering substance. Two tests on separate specimens can still share factors such as a patient’s disease severity, immune response, or a laboratory’s method. Repeating a test at a different time does not, by itself, establish independence.

Definition: Let $D$ denote the event that a person has the disease, and let $T_1+$ and $T_2+$ denote positive results from two tests. The results are conditionally independent given disease status if $$ P(T_1+ \cap T_2+ \mid D)=P(T_1+\mid D)P(T_2+\mid D) $$ and $$ P(T_1+ \cap T_2+ \mid D^c)=P(T_1+\mid D^c)P(T_2+\mid D^c). $$ Here $D^c$ means that the person does not have the disease. For Bayesian updating, these within-status joint probabilities—not merely the overall association between the tests—determine whether the likelihoods can be multiplied.

The same factorization condition applies to other combinations of results, such as positive then negative. In practice, to calculate the probability of a specified pair of results, the relevant joint probability must factor into the two individual probabilities within each disease-status group. If that condition does not hold, use the joint probability directly.

Bayesian Updating With Two Results

Write $E$ for the observed evidence—in this case, the pair of test results. Bayes’ rule says that the probability of disease after observing that evidence is

P(D\mid E)= \frac{P(E\mid D)P(D)} {P(E\mid D)P(D)+P(E\mid D^c)P(D^c)}.

The prior probabilities $P(D)$ and $P(D^c)$ describe disease status before these results are used. The quantities $P(E\mid D)$ and $P(E\mid D^c)$ are the likelihoods of the complete observed result pattern among people with and without the disease. For two positive results, if conditional independence holds, these likelihoods are

P(T_1+ \cap T_2+\mid D) =P(T_1+\mid D)P(T_2+\mid D), $$ $$ P(T_1+ \cap T_2+\mid D^c) =P(T_1+\mid D^c)P(T_2+\mid D^c).

Without conditional independence, those products must not be substituted for the joint likelihoods. The individual test sensitivities and false-positive rates alone do not determine the probability of a pair of results; the dependence between results matters as well.

Worked Example: A Shared Biological Factor

Consider a hypothetical screening test repeated twice on the same person. Among people with the disease, suppose half have a high level of a biological marker and half a low level. Given the marker level, assume the two test results are independent: the probability of a positive is 0.90 for the high-marker group and 0.20 for the low-marker group. Among people without the disease, suppose half have a high level of a technical interference factor and half a low level. Given that factor, the false-positive probabilities are 0.10 and 0.01, respectively. Assume the two results are independent given the relevant factor, but not necessarily given disease status alone. Let the disease prevalence be 0.01.

Likelihood of two positives among people with disease. Average over the two marker groups. Within each group, conditional independence allows the two positive probabilities to be multiplied:

P(T_1+ \cap T_2+\mid D) =0.50(0.90\times0.90)+0.50(0.20\times0.20) =0.405+0.020 =0.425.

Each test has marginal sensitivity $0.50(0.90)+0.50(0.20)=0.55$. Multiplying those marginal sensitivities would give $0.55^2=0.3025$, not the correct joint likelihood of 0.425.

Likelihood of two positives among people without disease.

P(T_1+ \cap T_2+\mid D^c) =0.50(0.10\times0.10)+0.50(0.01\times0.01) =0.005+0.00005 =0.00505.

Each test’s marginal false-positive probability is $0.50(0.10)+0.50(0.01)=0.055$. Its square is $0.003025$, again different from the joint likelihood.

Update the disease probability. Use the two joint likelihoods and the prior prevalence:

P(D\mid T_1+ \cap T_2+) =\frac{0.425(0.01)} {0.425(0.01)+0.00505(0.99)} =\frac{0.00425}{0.0092495} \approx 0.4595.

In this hypothetical setting, the probability of disease after two positive results is about 45.9%. The calculation using squared marginal rates would instead give $0.003025(0.01)/[0.003025(0.01)+0.003025(0.99)]$, which is 0.0100; that expression is not a valid update here because it uses the same false-positive probability for disease and non-disease groups. If the two marginal rates were used correctly in Bayes’ rule, the result would be $0.55^2(0.01)/[0.55^2(0.01)+0.055^2(0.99)]=0.003025/0.00601975\approx0.5025$. Either way, multiplying the marginal rates does not produce the correct posterior for this dependent-test model.

Why Shared Causes Change the Product

The example has an important structure. The results are independent once the shared factor is known, but they are dependent when that factor is unknown and only disease status is specified. The law of total probability shows how the dependence arises. Let $U$ represent the shared factor, with possible values $u$. If the tests are independent conditional on both $D$ and $U$, then

P(T_1+ \cap T_2+\mid D) =\sum_u P(T_1+ \cap T_2+\mid D,U=u)P(U=u\mid D) =\sum_u q_1(u)q_2(u)P(U=u\mid D),

where $q_i(u)=P(T_i+\mid D,U=u)$. In contrast, the product of the marginal positive probabilities is

P(T_1+\mid D)P(T_2+\mid D) =\left[\sum_u q_1(u)P(U=u\mid D)\right] \left[\sum_u q_2(u)P(U=u\mid D)\right].

These expressions are generally different: the first averages the products within factor groups, whereas the second multiplies averages across groups. When both tests are more likely to be positive in the same factor groups, the joint probability tends to exceed the product of the marginal probabilities. A person with a high-propensity factor is more likely to produce repeated positives, and a person with a low-propensity factor is more likely to produce repeated negatives. The shared factor makes matching results more common.

This is not the only possible pattern. Depending on the mechanism, results can also be negatively associated. The important point is not to assume a direction of dependence without evidence: use the joint probabilities appropriate to the testing process.

Conditional Independence and Ordinary Independence

Conditional independence given disease status is a different condition from ordinary, unconditional independence. Neither condition generally implies the other. This distinction matters because multiplying likelihoods for a Bayesian disease update requires conditional independence within the disease and non-disease groups, not ordinary independence in the overall population.

For instance, suppose half a population has the disease, and the tests are independent within each disease-status group. In the disease group, let each test be positive with probability 0.90; in the non-disease group, let each be positive with probability 0.10. Then each test is positive in half the population overall. But the overall probability of two positives is $0.50(0.90^2)+0.50(0.10^2)=0.41$, not $0.50\times0.50=0.25$. So the tests are conditionally independent given disease status but not ordinarily independent.

The reverse can happen too. Suppose $P(D)=P(D^c)=0.50$. Within the disease group, let the only possible pairs be $++$ and $--$, each with probability 0.50. Within the non-disease group, let the only possible pairs be $+-$ and $-+$, each with probability 0.50. In the overall population, each of the four pairs then has probability 0.25. Thus each test has positive probability 0.50 and the joint probability of two positives is 0.25, so the results are ordinarily independent. Yet within either disease-status group, the pair probabilities do not factor: for example, in the disease group, $P(++\mid D)=0.50$ while $P(T_1+\mid D)P(T_2+\mid D)=0.25$. Ordinary independence therefore does not license multiplying the within-group likelihoods.

Worked Example: Updating With Joint Probabilities

In a separate hypothetical setting, disease prevalence is 0.10. The tests have the following joint-result probabilities within each group:

Among people with disease, $P(T_1+)=0.80$, $P(T_2+)=0.80$, and $P(T_1+\cap T_2+)=0.68$.
Among people without disease, $P(T_1+)=0.05$, $P(T_2+)=0.05$, and $P(T_1+\cap T_2+)=0.04$.

These are valid joint probabilities. For example, among people with disease the remaining cells are $P(T_1+,T_2-)=0.12$, $P(T_1-,T_2+)=0.12$, and $P(T_1-,T_2-)=0.08$; all are nonnegative and the four cells sum to 1. Among people without disease, the remaining cells are 0.01, 0.01, and 0.94, also summing with 0.04 to 1.

Correct update after two positives. Use the supplied joint probabilities directly:

P(D\mid T_1+\cap T_2+) =\frac{0.68(0.10)} {0.68(0.10)+0.04(0.90)} =\frac{0.068}{0.104} \approx 0.6538.

Thus the posterior probability is about 65.4%. The conditional independence products would be $0.80^2=0.64$ and $0.05^2=0.0025$. Using those products instead would produce $0.064/(0.064+0.00225)\approx0.9660$, a much higher answer that is unsupported by the stated joint probabilities.

The sequential calculation makes the dependence especially clear. After the first positive, the disease probability is $0.08/(0.08+0.045)=0.64$. Given the first result, the probability of a second positive is $0.68/0.80=0.85$ for a person with disease and $0.04/0.05=0.80$ for a person without disease. The second result therefore changes the odds by a likelihood ratio of $0.85/0.80=1.0625$, not by the much larger marginal likelihood ratio $0.80/0.05=16$. Starting with odds $0.64/0.36=1.7778$, the updated odds are $1.7778(1.0625)=1.8889$, corresponding to probability $1.8889/(1+1.8889)\approx0.6538$.

A Practical Check Before Multiplying

Before using a product of test likelihoods, identify what independence claim is needed and whether it is plausible or supported. Ask whether the tests use the same specimen, assay method, equipment, reagent lot, or reader. Also consider shared features of the person or disease, including severity and biological response. These factors do not prove dependence in every setting, but they are reasons to seek joint repeat-test data or a model that represents the shared factor.

If a result is reported as a repeat-test performance rate, check what it conditions on. A rate measured across all people may not equal the probability of a second positive among people with the same disease status and the same first result. For sequential updating, the relevant quantity after observing $T_1+$ is $P(T_2+\mid T_1+,D)$ (and its counterpart given $D^c$). Under conditional independence given disease status, these equal the corresponding rates without conditioning on $T_1+$. Without that independence, use the conditional probabilities or joint probabilities that match the observed pattern.

Common mistake: “Each test has an 80% sensitivity, so two positive results have likelihood $0.80^2=0.64$ among people with disease.” This is correct only if the results are conditionally independent given disease status. If a shared biological or technical factor makes positive results cluster, the correct likelihood is $P(T_1+\cap T_2+\mid D)$, obtained from joint data or an appropriate model. Use the corresponding joint likelihood among people without disease as well before applying Bayes’ rule.

Key takeaway: Two repeated results are not automatically two independent pieces of evidence. For Bayesian updates, multiply individual likelihoods only when conditional independence holds within both disease-status groups; otherwise, update using the joint likelihood of the observed result pattern.

Check Your Understanding

Try these questions using the definitions and calculations above.

What independence condition is needed to write $P(T_1+\cap T_2+\mid D)$ as a product?
Give one biological and one technical factor that could make repeated test results dependent.
In the shared-factor example, why is the average of the within-group products not necessarily equal to the product of the marginal positive probabilities?
Can ordinary independence guarantee conditional independence given disease status? Explain briefly.
If joint probabilities for two positive results are available in the disease and non-disease groups, which probabilities should be used in Bayes’ rule?

Why Repeated Tests May Not Be Independent

What You'll Learn