MMP.16:5 - Archetypal Grounding

The cases use stipulated models and costs so that the design can be reconstructed. They demonstrate the method, not reports of performed experiments.

MMP.16:5.1 - The largest response difference disappears in the readout

A team has two candidate response laws for a calibrated device assumed valid over the input range 0 to 2:

• H1: y=x;
• H2: y=x^2.

Both agree at the previously inspected inputs 0 and 1. The team needs to know whether the modeled response at x=2 is below or above 3. The laws answer differently: 2 and 4. Their applicability across the range is an assumption supplied for this case.

With a numerical readout r=y+e and a known error bound -0.1 <= e <= 0.1, the design x=2 gives:

Now recover a missed feature of the actual instrument: it saturates at 1. Its record is r=min(1,y+e). At x=2 both alternatives always record 1. The large latent response difference gives no distinction at all.

Changing the input to x=0.5 yields:

This design separates the supplied alternatives without replacing the instrument. A record 0.27 retains H2; a record 0.52 retains H1. A record 0.37 fits neither under the given error bound, so it reopens the response or observing assumptions rather than forcing a label.

The result at x=0.5 supports the answer at x=2 through the supplied response families. If those families were justified only up to x=1, this design would not establish the requested extrapolation. The next work would concern that range extension or access to a suitable direct observation.

If the only available inputs were 0 and 1 and the record error law were the same under both accounts, the available designs would not separate them. That conclusion concerns the stated access; it does not say the response question is unanswerable under every possible instrument.

MMP.16:5.2 - A useful signal is still too costly

A service team must choose one of two recovery procedures. It models two possible failure modes H0 and H1 with current probabilities 0.5 each. The loss is remaining recovery time, in hours:

With current information, expected losses are 5 for A and 2 for B. The team chooses B.

A proposed diagnostic produces either “+” or “-”. Its supplied law is P(+ | H1)=0.8 and P(+ | H0)=0.2. Each signal has marginal probability 0.5. Bayes conditioning gives P(H1 | +)=0.8 and P(H1 | -)=0.2.

After “+”, A has expected loss 8 and B has 0.8, so the team chooses B. After “-”, A has expected loss 2 and B has 3.2, so it chooses A. Expected loss with the signal, excluding diagnostic cost, is therefore:

R(d) = 0.5*0.8 + 0.5*2 = 1.4 hours.
R0 - R(d) = 2 - 1.4 = 0.6 hours.

If obtaining, interpreting and waiting for the diagnostic adds 0.7 hours to recovery, its total expected loss is 2.1 hours; using current information is better under these conditions. At an all-inclusive cost of 0.2 hours, the diagnostic gives 1.6 hours and improves the choice. The example assumes the diagnostic does not change the failure mode or the two procedures.

A different test might reveal a device identifier perfectly. If that identifier is independent of the failure mode and losses in this model, its information gain about the identifier does not reduce this recovery loss. Selecting all available uncertainty as the target would misdirect the design.

Now change the current probability of H1 to 0.95. The “-” posterior is 0.19/(0.19+0.04), approximately 0.826, and the “+” posterior is 0.76/(0.76+0.01), approximately 0.987. B remains the better procedure after either result. The diagnostic still changes beliefs about the mode, but its sample-information value for this particular choice is zero.

A later method-development investigation could use the same signal for another target, such as learning how failure modes arise. That research value needs its own question and horizon; the zero above applies to the specified recovery choice.