A man tests for HIV. What is the predictive probability that his second test is negative?In a population, it is estimated HIV prevalence to be λ. For a new test for HIV:- θ is the probability of an HIV positive person to test positive- η is the probability an HIV negative person tests positive in this test.A person takes the test to check whether they have HIV, he tests positive.What is the predictive probability he tests negative on the second test?Assumption: Repeat tests on the same person are conditionally independent.From my notes predictive probability is given as: P ( Y ~ = y ~ | Y = y ) = ∫ p ( y ~ | τ ) p ( θ | τ ) here Y ~ is the unknown observable, y is the observed data and η the unknown.I am interested in the probability of the second test is negative, given that the first test is positive,without knowing if the man really has HIV or not.To facilitate this I define: y 1 as the event of the first test being positive and y 2 ~ as the second test being negativeWould this adaption to the formula given above be the correct/best approach to this problem ? p ( y 2 ~ , y 1 | τ ) = p ( y 2 ~ | τ ) p ( y 1 | τ ) p ( τ ) and this is really ∝ p ( y 2 ~ | τ ) p ( τ | y 1 )I've gotten for the p ( τ | y 1 ) from Bayes' theorem: p ( τ | y 1 ) = p ( τ ) p ( y 1 | τ ) p ( y 1 ) = λ θ λ θ + η ( 1 − λ ) How could I then find p ( y 2 ~ | τ )? Is this the correct approach?

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A man tests for HIV. What is the predictive probability that his second test is negative?In a population, it is estimated HIV prevalence to be   λ. For a new test for HIV:-   θ is the probability of an HIV positive person to test positive-   η is the probability an HIV negative person tests positive in this test.A person takes the test to check whether they have HIV, he tests positive.What is the predictive probability he tests negative on the second test?Assumption: Repeat tests on the same person are conditionally independent.From my notes predictive probability is given as:  P  (            Y      ~        =            y      ~            |    Y  =  y  )  =  ∫  p  (            y      ~            |    τ  )  p  (  θ      |    τ  ) here             Y      ~       is the unknown observable, y is the observed data and η the unknown.I am interested in the probability of the second test is negative, given that the first test is positive,without knowing if the man really has HIV or not.To facilitate this I define:      y    1   as the event of the first test being positive and                    y                  2                    ~       as the second test being negativeWould this adaption to the formula given above be the correct/best approach to this problem ?  p  (                    y                  2                    ~        ,      y          1            |    τ  )  =  p  (                    y                  2                    ~            |    τ  )  p  (      y          1            |    τ  )  p  (  τ  ) and this is really   ∝  p  (                    y                  2                    ~            |    τ  )  p  (  τ      |        y          1        )I&#039;ve gotten for the   p  (  τ      |        y          1        ) from Bayes&#039; theorem:  p  (  τ      |        y          1        )  =            p      (      τ      )      p      (              y        1                    |            τ      )              p      (              y        1            )          =            λ      θ              λ      θ      +      η      (      1      −      λ      )      How could I then find   p  (                    y                  2                    ~            |    τ  )? Is this the correct approach?

Kailee Abbott · Accepted Answer

Step 1I find it hard to follow your calculations, partly because you didn’t introduce   τ and your integrals don’t indicate their integration variables. Here’s one way to do this:                    P        (                  2nd test                       −                          ∣        1st test +        )                            =                              ∑                      σ            ∈            {            +            ,            −            }                          P        (                  2nd test                       −                          ∣        1st test +        ,        HIV        σ        )        P        (        HIV        σ        ∣        1st test +        )                                          =                              ∑                      σ            ∈            {            +            ,            −            }                          P        (                  2nd test                       −                          ∣        HIV        σ        )        P        (        HIV        σ        ∣        1st test +        )                                          =                              ∑                      σ            ∈            {            +            ,            −            }                          P        (                  2nd test                       −                          ∣        HIV        σ        )                              P            (            1st test +            ∣            HIV            σ            )            P            (            HIV            σ            )                                              ∑                              ρ                ∈                {                +                ,                −                }                                      P            (            1st test +            ∣            HIV            ρ            )            P            (            HIV            ρ            )                                                            =                                                        ∑                              σ                ∈                {                +                ,                −                }                                      P            (                          2nd test                               −                                      ∣            HIV            σ            )            P            (            1st test +            ∣            HIV            σ            )            P            (            HIV            σ            )                                              ∑                              ρ                ∈                {                +                ,                −                }                                      P            (            1st test +            ∣            HIV            ρ            )            P            (            HIV            ρ            )                                  ,            where the first equality applies the law of total probability, the second equality applies your assumption of conditional independence of multiple tests, the third equality applies Bayes’ theorem to express   P  (  HIV  μ  ∣  1st test +  ) in terms of known quantities, and the fourth equality is just a rearrangement of the sum.Step 2Another way to get the same result is by applying the law of total probability to both the numerator and the denominator in  P  (      2nd test           −        ∣  1st test +  )  =            P      (              2nd test                   −                    ∩      1st test +      )              P      (      1st test +      )          .Plugging in your variables yields                    P        (                  2nd test                       −                          ∣        1st test +        )                            =                                          (            1            −            θ            )            θ            λ            +            (            1            −            η            )            η            (            1            −            λ            )                                θ            λ            +            η            (            1            −            λ            )                                  .

A man tests for HIV. What is the predictive probability that his second test is negative?

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