# Independence of diagnostic tests

A rare disease afflicts only $$1\%$$ of the population. Fred gets tested for it with two diagnostic tests. The sensitivity and specificity of the second test are both $$0.95$$ (so, its overall accuracy is $$0.95$$), and knowing that Fred has the disease makes the second test independent of the first.

Let $$D$$ be the event that Fred has disease, $$T_1$$ be the event that his first result is positive and $$T_2$$ be the event that his second result is positive.

I would like to know if I am interpreting each of the following correctly:

1. $$P(T_1 \cap T_2 | D)$$ : Probability that the results are positive in both the tests given Fred has the disease.

2. $$P(T_1 \cap T_2)$$ : Probability that the results are positive in both the tests taking into account both the cases - when Fred has the disease and when not.

Does this mean that $$P(T_1 \cap T_2) = P(T_1 \cap T_2 | D) P(D) + P(T_1 \cap T_2 | D^c) P(D^c)$$?

3. Is it true that $$P(T_1 \cap T_2 | D) = P(T_1 | D) P(T_2 | D)$$?

4. Is it true that $$P(T_1 \cap T_2) = P(T_1)P(T_2)$$?

My computations show that $$(3)$$ holds while $$(4)$$ does not, but I can't intuit why; an explanation using (conditional) independence would be helpful.

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1. $$P(T_1 \cap T_2 | D)$$ : Probability that the results are positive in both the tests given Fred has the disease.

2. $$P(T_1 \cap T_2)$$ : Probability that the results are positive in both the tests taking into account both the cases - when Fred has the disease and when not.

Does this mean that $$P(T_1 \cap T_2) = P(T_1 \cap T_2 | D) P(D) + P(T_1 \cap T_2 | D^c) P(D^c)$$?

Yes, yes, and yes.

A rare disease afflicts only $$1\%$$ of the population. Fred gets tested for it with two diagnostic tests. The sensitivity and specificity of the second test are both $$0.95,$$ and knowing that Fred has the disease makes the second test independent of the first.

1. Is it true that $$P(T_1 \cap T_2 | D) = P(T_1 | D)\times P(T_2 | D)$$?

This is definitionally equivalent to the statement “events $$T_1$$ and $$T_2$$ are conditionally independent given event $$D.$$

It is true here, since knowing that Fred has the disease makes the two tests independent. The computation: $$P(T_1 T_2 \mid D) = 0.95x = (0.95x +0.05x)\times0.95= P(T_1 \mid D)\times P(T_2 \mid D).$$

1. Is it true that $$P(T_1 \cap T_2) = P(T_1)\times P(T_2)$$?

Not necessarily, even in the given scenario: $$P(T_1 T_2) \\=(0.01)(0.95)x+(0.99)(0.05)y \\=0.0095x+0.0495y \\\not\equiv 0.00059x+0.0029205y+0.0554895z \\=\Big(0.01x+(0.99)(0.05y+0.95z)\Big) \times \Big((0.01)(0.95)+(0.99)(0.05)\Big) \\=P(T_1)\times P(T_2).$$

So, the two tests being independent conditional on Fred having the disease does not imply that the two tests are independent! ҂ †

but aren't medical tests supposed to be independent regardless of the disease status?

No, medical tests are not in general independent of each other! For example, if two COVID swab tests are performed in succession on the same subject, the second test is less likely than the first to register positive, due a lower concentration of antigens in the second reagent tube.

҂ More interestingly, the converse is also false: independence does not imply conditional independence (so, neither independence nor conditional independence implies the other).

In the following example, $$T_1$$ and $$T_2$$ are independent, yet, given that $$D$$ happens, $$T_1$$ and $$T_2$$ are dependent!

$$P(T_1T_2)=\frac14=P(T_1)\times P(T_2),$$ yet, given that $$\boldsymbol D$$ happens, $$P(T_1 T_2) =\frac13\ne \frac49 = P(T_1)\times P(T_2).$$

Notably, given that $$\boldsymbol{D^\complement}$$ happens, $$P(T_1 T_2) = 0 = P(T_1)\times P(T_2);$$ hence, $$T_1$$ and $$T_2$$ being either dependent or independent conditional on whether $$D$$ happens does not imply that $$T_1$$ and $$T_2$$ are dependent!!

Remarkably, events $$X$$ and $$Y$$ being independent conditional on event $$A$$ as well as conditional on event $$A^\complement$$ still does not imply that $$X$$ and $$Y$$ are independent!!

For example, let $$E$$ be an event with probability strictly between $$0$$ and $$1;$$ then $$P(E\cap E\mid E)=1=P(E\mid E)\times P(E\mid E),\\ P(E\cap E\mid E^\complement)=0=P(E\mid E^\complement)\times P(E\mid E^\complement),$$ yet $$P(E\cap E)=P(E)\ne [P(E)]^2=P(E)\times P(E).$$

Alternatively, here's a less abstract, coin-toss example.

• Interestingly, during calculations (when I match my answer to the original problem), I find that $P(T_1 T_2 | D) = P(T_1 | D) P(T_2 | D)$ and $P(T_1 T_2) \neq P(T_1) P(T_2)$. I think it's because the statement given says "The new test is independent of the first test (given his disease status)", but aren't medical tests supposed to be independent regardless of the disease status? Jun 23 at 20:52
• @fractalletter On closer inspection, I've revised my answer. Hope you enjoy reading it! 2 days ago
• Hi, thank you for the detailed, illustrative answer! One final question: How can "the tests be independent given Fred has the disease" but not otherwise (when Fred doesn't have it)? I know it follows from the given statement, but can you pick an example from reality (like you did with Covid swab)? Thanks! 2 days ago
• @fractalletter I expanded the answer to make it more complete. Re: your question, both the probability tree & Venn-diagram examples above are such that $T_1$ and $T_2$ are either dependent or independent conditional on whether $D$ happens; just take the abstract Venn diagram, let the sample space be four playing cards, and $T_1,$ $T_2$ and $D$ correspond to star, blue, and digit 7. 2 days ago
• I see, thank you once again! :) yesterday