ENGI 3423 Probability and Statistics

Extension of Bayes’ Theorem to a Pair of Tests

Suppose that an input signal is one of the mutually exclusive and collectively exhaustive set of signals { A₁, A₂, A₃, ... , A_n } and that a test is available to determine which signal was input.   However, the test is not perfect.   It is possible that the test will claim that a signal other than the true one was input.   The same test is applied twice to the same input signal.

Let   B_j   represent the claim from the first application of the test and
let   C_k   represent the claim from the second application of the test.

The probabilities   P[A_i],   P[B_j | A_i ]   and   P[C_k | A_i B_j ]   are all known.
Furthermore, we can assume that the results of the two tests are independent of each other, so that
P[C_k | A_i B_j ] = P[C_k | A_i ]   and we can also assume that the two tests are identical, so that
P[C_j | A_i ] = P[B_j | A_i ].

Then after the first application of the test, Bayes’ theorem tells us that

Extended to the pair of tests, this becomes (see the proof below)

But, because the tests are independent and identical (and therefore exchangeable),

which is the same as the single test result, except that the posterior probabilities from the first test
{ P[ A_i | B_j] } replace { P[ A_i ] } as the prior probabilities in the second test.

An application of this result to a system of three signals, { A₁, A₂, A₃ } and a pair of tests is provided in an associated Excel file.

Proof: