A Multi-Stage Diagnostic And Claim Calculation

Why the Conditioning Order Matters

A diagnostic-and-claim problem can require several probability steps before any payment is calculated. First, the test result changes the probability of the underlying condition. Next, the loss distribution depends on whether that condition is present. Finally, a deductible changes each loss into a payment. Skipping or reversing one of these steps can produce a plausible-looking but incorrect answer.

The essential discipline is to keep the events distinct. Let $D$ denote the event that a condition is present, and let $T^+$ denote a positive test result. The test’s sensitivity is $P(T^+\mid D)$, while its false-positive probability is $P(T^+\mid D^c)$. Neither is the probability of the condition given a positive result, $P(D\mid T^+)$. Bayes’ theorem supplies that posterior probability.

Definition: For a loss $X$ subject to a per-loss deductible $d$, with no policy limit or coinsurance, the insurer’s payment per loss is $Y^L=(X-d)_+=\max(X-d,0)$. If the loss distribution depends on a condition $D$, calculate the payment under each conditional distribution before combining the results.

Unless a problem specifies otherwise, the calculations below concern undiscounted amounts at the time of loss. No interest rate or mortality assumption is involved. The examples explicitly state their diagnostic and loss-model assumptions.

A Reliable Calculation Sequence

For a positive result, Bayes’ theorem gives

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

For a negative result, use $P(T^-\mid D)$ and $P(T^-\mid D^c)$ in the same formula. The denominator must match the result being conditioned on: it is the total probability of that test result.

Once the posterior probability is found, use the law of total expectation. If the loss distribution conditional on $D$ is $X\mid D$, and conditional on $D^c$ is $X\mid D^c$, then

E[Y^L\mid T^+] = P(D\mid T^+)E[(X-d)_+\mid D,T^+] + P(D^c\mid T^+)E[(X-d)_+\mid D^c,T^+].

If the diagnostic result and the loss are conditionally independent given whether $D$ is present, then the loss distribution given both $D$ and $T^+$ is just the loss distribution given $D$. The analogous statement applies under $D^c$. This conditional-independence assumption is important: if the test result carries additional information about claim severity even after the condition is known, the loss distributions must be conditioned on the test result as well.

How the Deductible Changes an Expected Loss

For any nonnegative loss $X$, a deductible does not simply subtract $d$ from the expected loss. Losses at or below the deductible produce no payment. For a continuous loss with density $f_X$, the expected payment is

E[(X-d)_+] = \int_d^\infty (x-d)f_X(x)\,dx.

This formula follows directly from the payment definition: the payment is zero for $x\le d$, and it is $x-d$ for $x>d$. Equivalently, for a nonnegative loss with survival function $S_X(x)=P(X>x)$,

E[(X-d)_+] = \int_d^\infty S_X(x)\,dx.

To see why, write each nonnegative payment $y$ as $\int_0^\infty \mathbf{1}_{\{y>t\}}\,dt$, where $\mathbf{1}$ is the indicator of an event. Taking expectations gives $E[Y]=\int_0^\infty P(Y>t)\,dt$. For $Y=(X-d)_+$, the event $Y>t$ is $X>d+t$; changing the integration variable gives the displayed survival-function formula. For a discrete loss, instead sum each possible payment multiplied by its probability.

Worked Example: A Positive Test and Two Exponential Loss Models

In a hypothetical screening program, 2% of people have condition $D$. A test has sensitivity 0.90 and specificity 0.95, so $P(T^+\mid D)=0.90$ and $P(T^+\mid D^c)=0.05$. Assume the test result and claim loss are conditionally independent given $D$. Conditional on $D$, the loss $X$ is exponential with mean $4,000; conditional on $D^c$, it is exponential with mean $500. A policy applies a $1,000 deductible. Find the expected insurer payment per loss given a positive result.

Method. Update the condition probability first, then calculate the deductible payment under each exponential model, and finally weight those payments by the posterior probabilities. There is no interest or mortality model; amounts are undiscounted dollars.

1. Posterior probability. The probability of a positive result is $0.90(0.02)+0.05(0.98)=0.018+0.049=0.067$. Therefore,

P(D\mid T^+)=\frac{0.018}{0.067}=\frac{18}{67}, \qquad P(D^c\mid T^+)=\frac{49}{67}.

2. Payment under $D$. If $X$ is exponential with mean $\theta$, its survival function is $e^{-x/\theta}$, and the deductible expectation is $\theta e^{-d/\theta}$. Here the mean is $4,000 and the deductible is $1,000:

E[(X-1000)_+\mid D] = 4000e^{-1000/4000} = 4000e^{-0.25} \approx \$3{,}115.20.

3. Payment under $D^c$. The conditional mean is $500, so

E[(X-1000)_+\mid D^c] = 500e^{-1000/500} = 500e^{-2} \approx \$67.67.

4. Combine. Using the posterior weights,

E[Y^L\mid T^+] = \frac{18}{67}(3115.203\ldots) + \frac{49}{67}(67.6676\ldots) \approx \$886.41.

The expected payment is $886.41 per loss given a positive test, rounded to the nearest cent. The final value is between the two conditional expected payments, as a weighted average must be.

Worked Example: A Discrete Loss Model and a Deductible

In a hypothetical equipment-monitoring program, 10% of units have a fault $D$. A monitor has sensitivity 0.80 and specificity 0.90. Assume its result and the loss are conditionally independent given fault status. Given $D$, the loss $X$ is $0 with probability 0.20, $2,000 with probability 0.50, and $5,000 with probability 0.30. Given $D^c$, it is $0 with probability 0.50, $1,200 with probability 0.30, and $3,000 with probability 0.20. Find the expected payment given a positive monitor result under a $1,000 deductible.

Method. Use Bayes’ theorem for the positive result and sum the deductible-adjusted payments over the loss values. No interest or mortality assumption is used; all amounts are undiscounted dollars.

The probability of a positive result is $0.80(0.10)+0.10(0.90)=0.08+0.09=0.17$. Thus

P(D\mid T^+)=\frac{0.08}{0.17}=\frac{8}{17}, \qquad P(D^c\mid T^+)=\frac{9}{17}.

Under $D$, the three possible insurer payments are $\max(0-1000,0)=0$, $\max(2000-1000,0)=1000$, and $\max(5000-1000,0)=4000$. Therefore,

E[Y^L\mid D] = 0.20(0)+0.50(1000)+0.30(4000) = \$1{,}700.

Under $D^c$, the payments are $0, $200, and $2,000. Hence

E[Y^L\mid D^c] = 0.50(0)+0.30(200)+0.20(2000) = \$460.

Combining the conditional expectations gives

E[Y^L\mid T^+] = \frac{8}{17}(1700)+\frac{9}{17}(460) = \frac{17740}{17} \approx \$1{,}043.53.

The result is $1,043.53 per loss given a positive result, rounded to the nearest cent. A useful check is that this lies between $460 and $1,700.

When Conditional Loss Models Have Different Shapes

The method does not require the two conditional loss distributions to have the same form. One might be shifted exponential and the other uniform, for example. The essential requirements are to apply the deductible to each model and to use the posterior probabilities for the particular test result.

Worked Example: A Shifted Exponential and a Uniform Loss

In a hypothetical device-protection plan, 10% of devices have a defect $D$. A diagnostic test has sensitivity 0.90 and specificity 0.95. Assume conditional independence of test result and loss given defect status. Given $D$, let $X=600+Z$, where $Z$ is exponential with mean $2,000. Given $D^c$, let $X$ be uniform on $0 to $3,000. Find the expected payment given a positive result with a $1,000 deductible.

Method. Find the positive-result posterior, then account for where the deductible falls within each loss distribution. There is no interest or mortality assumption; amounts are undiscounted dollars.

The probability of a positive result is $0.90(0.10)+0.05(0.90)=0.09+0.045=0.135$, so

P(D\mid T^+)=\frac{0.09}{0.135}=\frac{2}{3}, \qquad P(D^c\mid T^+)=\frac{1}{3}.

Under $D$, the payment is $(600+Z-1000)_+=(Z-400)_+$. Applying the exponential deductible formula to $Z$, whose mean is $2,000, gives

E[Y^L\mid D] = 2000e^{-400/2000} = 2000e^{-0.2} \approx \$1{,}637.46.

Under $D^c$, $X$ is uniform on $[0,3000]$. The payment is positive only when $X>1000$, so its expectation is the area under the payment function divided by the interval length:

E[Y^L\mid D^c] = \frac{1}{3000}\int_{1000}^{3000}(x-1000)\,dx = \frac{(3000-1000)^2}{2(3000)} = \$666.67.

Therefore,

E[Y^L\mid T^+] = \frac{2}{3}(1637.4615\ldots) + \frac{1}{3}(666.6667\ldots) \approx \$1{,}313.86.

The expected payment is $1,313.86 per loss, rounded to the nearest cent. The uniform calculation uses only losses above the deductible; subtracting $1,000 from the unconditional mean would not represent this payment rule.

Common Traps and Checks

A frequent error is to use the prevalence $P(D)$ as the weight after observing a test result. In the first example, using 0.02 instead of $P(D\mid T^+)=18/67$ would give

0.02(3115.203\ldots)+0.98(67.6676\ldots) \approx \$128.62,

not $886.41. That incorrect calculation ignores the information in the positive test. Another common error is to subtract the deductible from the mean loss. In the discrete example, $E[X\mid D]=0.20(0)+0.50(2000)+0.30(5000)=\$2{,}500$; subtracting $1,000 gives $1,500, whereas the correct expected payment is $1,700. The loss of $0 produces no payment and cannot contribute a negative amount.

For a positive result, use positive-result likelihoods in both the numerator and denominator of Bayes’ formula. For a negative result, use negative-result likelihoods.
Apply $(X-d)_+$ to the loss, not $X-d$ without a zero floor.
Do not replace conditional loss models with one unconditional model before applying the deductible.
Check that posterior weights are nonnegative and sum to 1, and that a weighted expected payment lies between its conditional expected payments.

These steps keep the diagnostic update, the conditional loss model, and the contract payment rule separate until the final combination. This structured approach is aligned with the SOA Exam P (Probability) syllabus and is useful whenever evidence changes the relevant risk mix.

Check Your Understanding

For each problem, calculate the posterior condition probability first, then apply the deductible within each conditional loss model.

A condition has prevalence 0.04. A test has sensitivity 0.75 and specificity 0.90. Given the condition, losses are $0, $2,000, and $6,000 with probabilities 0.25, 0.50, and 0.25. Without the condition, losses are $0 and $1,500 with probabilities 0.70 and 0.30. Assume conditional independence of test result and loss given condition status. What is expected payment given a positive test with a $1,000 deductible? (Answer: $530.95)
A condition has prevalence 0.08. A test has sensitivity 0.90 and specificity 0.95. Conditional on condition status, the test result and loss are independent. Given the condition, loss is exponential with mean $3,000; without it, loss is exponential with mean $600. Find expected payment given a negative result with a $900 deductible. (Answer: $152.82)
A condition has prevalence 0.20. A test has sensitivity 0.85 and specificity 0.90. Given the condition, loss is uniform on $500 to $4,500; without it, loss is uniform on $0 to $2,000. Assume conditional independence of test result and loss given condition status. Find expected payment given a positive test with a $1,500 deductible. (Answer: $785.00)
A condition has prevalence 0.05. A test has sensitivity 0.80 and specificity 0.95. Given the condition, loss is $800, $1,800, or $4,000 with probabilities 0.20, 0.50, and 0.30. Without the condition, loss is $0 or $1,200 with probabilities 0.60 and 0.40. Assume conditional independence of test result and loss given condition status. Find expected payment given a positive result with a $1,000 deductible. (Answer: $637.71)

What You'll Learn