I am running into a problem with two tests run serially.
| Sensitivity | Specificity | |
|---|---|---|
| Test 1 | 0.95 | 0.90 |
| Test 2 | 0.94 | 0.89 |
Tests are run serially, thus both tests must pass in order for the test battery to pass, and only one of the two (or both) tests must fail in order for the battery to fail.
Tests are conditionally independent on event A, which represents having a disease.
I am interested in calculating the Sensitivity of the entire test battery, as well as the Specificity of the entire test battery.
The Sensitivity is pretty easy:
Let $H_1$ = Test 1 is positive,
Let $H_2$ = Test 2 is positive, and
Let $A$ = Having the disease
Given that the tests are conditionally independent given A, we can say:
$P(H_1 \cap H_2|A) = P(H_1|A)P(H_2|A)$
= 0.95 * 0.94
= 0.8930 = Sensitivity
I run into a problem with specificity, however:
Since the tests are run serially, one of the two, or both, need to fail. So we want to find:
$P(H_1^c \cup H_2^c |A^c)$
This equals $P( (H_1 \cap H_2)^c |A^c)$.
since $P(A^c|B) = 1 – P(A|B)$,
I can say that
$P( (H_1 \cap H_2)^c |A^c) = 1- P(H_1 \cap H_2 | A^c)$
This is where I run into my confusion. From https://en.wikipedia.org/wiki/Conditional_independence, I know that
$P(H_1 \cap H_2 |A) = P(H_1|A)P(H_2|A)$, however:
$P(H_1 \cap H_2 |A^c) \neq P(H_1|A^c)P(H_2|A^c)$
So then how can I prove that the Specificity of the test battery is in fact equal to
$1 – P(H_1|A^c)P(H_2|A^c)$ since this requires that $1- P(H_1 \cap H_2 | A^c) = 1 – P(H_1|A^c)P(H_2|A^c)$ ???
note that my textbook states:
$Spec = 1 − [(1 − Spec_1) × (1 − Spec_2) × ··· × (1 − Sp_k)]$ which i believe is the same as $1 – P(H_1|A^c)P(H_2|A^c)$ in the case of two tests
thank you
Best Answer
I believe your source has a typo and should say that the tests are independent conditional on disease status (i.e. $A$ and also $A^c$). You have already pointed out the mathematical details, so I shall furnish you with some hastily obtained sources to bolster your confidence that this more complete assumption is indeed typically what is meant when "conditional independence" is assumed in multiple tests:
-See this calculator: "Calculations assume that the two tests are independent, conditional on disease status"
-See p. 10 and Rmk 2 on p. 4 here : "...it is reasonable to assume that $T_1^+$ and $T_2^+$ are conditionally independent under $D$ as well as under $D′$ "
-See p. 148 here for their specification of the "conditional independence" assumption :"The model assumes that the two test outcomes for a given animal are independent (conditional on the infection status of the animal)..." and see p. 159 for how your textbook's formulae for two tests are justified under that specification