Dual processing model for medical decision-making:
An extension to diagnostic testing
Appendix
Derivation of the testing thresholds
Considering the decision tree in Fig. 2 we have:
Decision Weight Outcome Utility
Valuation
Probability
𝑥1
𝑈𝐼,1
𝑥1 𝐼 = 𝑅𝑔(𝑥1 ) = 0
𝛾
𝑥2
𝑈𝐼,2
𝑥2 𝐼 = 𝑅𝑔(𝑥2 ) = 𝑈𝐼,4 − 𝑈𝐼,2
(1 − 𝛾)
𝑥1
𝑈𝐼𝐼,1
𝑥1
(1 − 𝛾)
𝑥2
𝑈𝐼𝐼,2
𝑥2
𝑅𝑥
1
2
1
2
𝑚
𝛾
𝑚
𝑚𝐼𝐼
= 𝑈𝐼𝐼,1
𝑝
𝑚𝐼𝐼
= 𝑈𝐼𝐼,2
1−𝑝
Note that under Type I processes, every outcome with non-zero probability is assigned
equal weight. Therefore, in the two alternative outcomes case each probability is
considered equal to 0.5 [12, 53]. The expected valuation of the decision to treat (Rx) is
computed as the summation of the valuations of each outcome:
𝛾
𝑉(𝑅𝑥) = (𝑈𝐼,2 − 𝑈𝐼,4 ) + (1 − 𝛾)[𝑝𝑈𝐼𝐼,1 + (1 − 𝑝)𝑈𝐼𝐼,2 ]
2
Decision Weight Outcome Utility
Valuation
Probability
𝑚
𝛾
𝑥3
𝑈𝐼,3
𝑥3 𝐼 = 𝑅𝑔(𝑥3 ) = 𝑈𝐼,1 − 𝑈𝐼,3
𝛾
𝑥4
𝑈𝐼,4
𝑥4 𝐼 = 𝑅𝑔(𝑥4 ) = 0
(1 − 𝛾)
𝑥3
𝑈𝐼𝐼,3
𝑥3
(1 − 𝛾)
𝑥4
𝑈𝐼𝐼,4
𝑥4
𝑁𝑜𝑅𝑥
𝑚
1
2
1
2
𝑚𝐼𝐼
= 𝑈𝐼𝐼,3
𝑝
𝑚𝐼𝐼
= 𝑈𝐼𝐼,4
1−𝑝
The expected valuation of the decision not to treat (NoRx) is computed as the
summation of the valuations of each outcome:
1
𝑉(𝑁𝑜𝑅𝑥) =
Decision
𝑇
𝛾
(𝑈 − 𝑈𝐼,1 ) + (1 − 𝛾)[𝑝𝑈𝐼𝐼,3 + (1 − 𝑝)𝑈𝐼𝐼,4 ]
2 𝐼,3
Weight
Outcome
Utility
Valuation
Probability
𝛾
𝑥1𝑇
𝑈𝐼,1 − 𝐻𝐼,𝑇
𝑥1𝑇𝐼 = 𝑅𝑔(𝑥1𝑇 ) = −𝐻𝐼,𝑇
𝛾
𝑥2𝑇
𝑈𝐼,2 − 𝐻𝐼,𝑇
𝑥2𝑇𝐼 = 𝑅𝑔(𝑥2𝑇 ) = 𝑈𝐼,2 − 𝑈𝐼,4 − 𝐻𝐼,𝑇
𝛾
𝑥3𝑇
𝑈𝐼,3 − 𝐻𝐼,𝑇
𝑥3𝑇𝐼 = 𝑅𝑔(𝑥3𝑇 ) = 𝑈𝐼,3 − 𝑈𝐼,1 − 𝐻𝐼,𝑇
𝛾
𝑥4𝑇
𝑈𝐼,4 − 𝐻𝐼,𝑇
𝑥4𝑇𝐼 = 𝑅𝑔(𝑥4𝑇 ) = −𝐻𝐼,𝑇
(1 − 𝛾)
𝑥1𝑇
𝑈𝐼𝐼,1 − 𝐻𝐼𝐼,𝑇
𝑥1𝑇𝐼𝐼 = 𝑈𝐼𝐼,1 − 𝐻𝐼𝐼,𝑇
(1 − 𝛾)
𝑥2𝑇
𝑈𝐼𝐼,2 − 𝐻𝐼𝐼,𝑇
𝑥2𝑇𝐼𝐼 = 𝑈𝐼𝐼,2 − 𝐻𝐼𝐼,𝑇
(1 − 𝛾)
𝑥3𝑇
𝑈𝐼𝐼,3 − 𝐻𝐼𝐼,𝑇
𝑥3𝑇𝐼𝐼 = 𝑈𝐼𝐼,3 − 𝐻𝐼𝐼,𝑇
(1 − 𝛾)
𝑥4𝑇
𝑈𝐼𝐼,4 − 𝐻𝐼𝐼,𝑇
𝑥4𝑇𝐼𝐼 = 𝑈𝐼𝐼,4 − 𝐻𝐼𝐼,𝑇
1
4
1
4
1
4
1
4
𝑚
𝑚
𝑚
𝑚
𝑚
𝑝𝑆
𝑚
(1 − 𝑝)(1 − 𝑆𝑝 )
𝑚
𝑝(1 − 𝑆)
𝑚
(1 − 𝑝)𝑆𝑝
Note that under Type I processes, every outcome with non-zero probability is assigned
equal weight. Therefore, in the four alternative outcomes case each probability is
considered equal to 0.25 [12, 53]. The expected valuation of the decision to test (T) is
computed as the summation of the valuations for each outcome:
𝛾
𝑉(𝑇) = (𝑈𝐼,2 − 𝑈𝐼,4 + 𝑈𝐼,3 − 𝑈𝐼,1 )
4
+ (1 − 𝛾)[𝑝𝑆𝑈𝐼𝐼,1 + (1 − 𝑝)(1 − 𝑆𝑝 )𝑈𝐼𝐼,2 + 𝑝(1 − 𝑆)𝑈𝐼𝐼,3 + (1 − 𝑝)𝑆𝑝 𝑈𝐼𝐼,4 ]
− (𝛾𝐻𝐼,𝑇 + (1 − 𝛾)𝐻𝐼𝐼,𝑇 )
Thresholds
We define the benefits and harms under type 1 and 2 as follows: 𝐵𝐼 = 𝑈𝐼,1 − 𝑈𝐼,3, 𝐻𝐼 =
𝑈𝐼,4 − 𝑈𝐼,2 and 𝐵𝐼𝐼 = 𝑈𝐼𝐼,1 − 𝑈𝐼𝐼,3, 𝐻𝐼𝐼 = 𝑈𝐼𝐼,4 − 𝑈𝐼𝐼,2. We assume that and 𝐻𝐼 , 𝐻𝐼𝐼 > 0
Testing threshold
We set 𝑉(𝑁𝑜𝑅𝑥) = 𝑉(𝑇), which results in:
𝛾
𝛾
(𝑈𝐼,3 − 𝑈𝐼,1 ) + (1 − 𝛾)[𝑝𝑈𝐼𝐼,3 + (1 − 𝑝)𝑈𝐼𝐼,4 ] = (𝑈𝐼,2 − 𝑈𝐼,4 + 𝑈𝐼,3 − 𝑈𝐼,1 )
2
4
2
+(1 − 𝛾)[𝑝𝑆𝑈𝐼𝐼,1 + (1 − 𝑝)(1 − 𝑆𝑝 )𝑈𝐼𝐼,2 + 𝑝(1 − 𝑆)𝑈𝐼𝐼,3 + (1 − 𝑝)𝑆𝑝 𝑈𝐼𝐼,4 ]
− (𝛾𝐻𝐼,𝑇 + (1 − 𝛾)𝐻𝐼𝐼,𝑇 )
𝛾
(𝐻 − 𝐵𝐼 ) + (1 − 𝛾)[𝑝𝑈𝐼𝐼,3 + (1 − 𝑝)𝑈𝐼𝐼,4 ] + (𝛾𝐻𝐼,𝑇 + (1 − 𝛾)𝐻𝐼𝐼,𝑇 )
4 𝐼
= (1 − 𝛾)[𝑝𝑆𝑈𝐼𝐼,1 + (1 − 𝑝)(1 − 𝑆𝑝 )𝑈𝐼𝐼,2 + 𝑝(1 − 𝑆)𝑈𝐼𝐼,3 + (1 − 𝑝)𝑆𝑝 𝑈𝐼𝐼,4 ]
𝛾
(1 − 𝛾)(1 − 𝑆𝑝 )𝐻𝐼𝐼 + (1 − 𝛾)𝐻𝐼𝐼𝑇 + (𝐻𝐼 − 𝐵𝐼 ) + 𝛾𝐻𝐼,𝑇 = (1 − 𝛾)𝑝[𝑆𝐵𝐼𝐼 + (1 − 𝑆𝑝 )𝐻𝐼𝐼 ]
4
(1 − 𝑆𝑝 )𝐻𝐼𝐼 + 𝐻𝐼𝐼,𝑇
𝑆𝐵𝐼𝐼 + (1 − 𝑆𝑝 )𝐻𝐼𝐼
+
(𝐻𝐼 − 𝐵𝐼 )
𝛾
1
𝛾𝐻𝐼,𝑇
+
=𝑝
4(1 − 𝛾) 𝑆𝐵𝐼𝐼 + (1 − 𝑆𝑝 )𝐻𝐼𝐼 (1 − 𝛾) 𝑆𝐵𝐼𝐼 + (1 − 𝑆𝑝 )𝐻𝐼𝐼
The formula for the threshold between withholding the treatment and testing becomes:
1 𝐻
1 + 1 − 𝑆 𝐻𝐼𝐼,𝑇
𝛾
𝐵𝐼 − 𝐻𝐼
1
𝛾𝐻𝐼,𝑇
𝑝
𝐼𝐼
𝑝𝑡𝑡 =
−
+
𝑆 𝐵
4(1 − 𝛾) 𝑆𝐵𝐼𝐼 + (1 − 𝑆𝑝 )𝐻𝐼𝐼 (1 − 𝛾) 𝑆𝐵𝐼𝐼 + (1 − 𝑆𝑝 )𝐻𝐼𝐼
1 + 1 − 𝑆 𝐻𝐼𝐼
𝑝 𝐼𝐼
1 𝐻
1 + 1 − 𝑆 𝐻𝐼𝐼,𝑇
1
𝐻𝐼 − 𝐵𝐼
𝛾𝐻𝐼,𝑇
𝑝
𝐼𝐼
𝑝𝑡𝑡 =
+
(𝛾
+4
)
𝑆 𝐵
𝑆 𝐵
(1 − 𝑆𝑝 )𝐻𝐼𝐼
1 + 1 − 𝑆 𝐻𝐼𝐼
4(1 − 𝛾) (1 + 1 − 𝑆 𝐻𝐼𝐼 ) (1 − 𝑆𝑝 )𝐻𝐼𝐼
𝑝 𝐼𝐼
𝑝 𝐼𝐼
1 𝐻𝐼𝐼,𝑇
1 − 𝑆𝑝 𝐻𝐼𝐼
𝛾
𝐻𝐼
𝐵𝐼
𝐻𝐼,𝑇
𝑝𝑡𝑡 =
1+
( (1 − ) + 4
)
𝑆 𝐵
1 𝐻𝐼𝐼,𝑇 𝐻𝐼𝐼
𝐻𝐼
𝐻𝐼𝐼
1 + 1 − 𝑆 𝐻𝐼𝐼
4(1 − 𝛾)(1 − 𝑆𝑝 ) (1 + 1 − 𝑆 𝐻 )
𝑝 𝐼𝐼 [
]
𝑝
𝐼𝐼
1+
𝑝𝑡𝑡 = 𝑝𝑡𝑡 (𝐸𝑈𝑇) 1 +
[
𝛾
𝐻𝐼
𝐵𝐼
𝐻𝐼,𝑇
( (1 − ) + 4
)
1 𝐻
𝐻
𝐻𝐼
𝐻𝐼𝐼
4(1 − 𝛾)(1 − 𝑆𝑝 ) (1 + 1 − 𝑆 𝐻𝐼𝐼,𝑇 ) 𝐼𝐼
]
𝑝
𝐼𝐼
To constrain the values of 𝑝𝑡𝑡 ∈ [0,1], we modify the 𝑝𝑡𝑡 formula such as:
𝑝𝑡𝑡 = min {𝑝𝑡𝑡,𝐸𝑈𝑇 1
[
+
𝛾
𝐻𝐼
𝐵𝐼
𝐻𝐼,𝑇
(1 − ) + 4
) , 1} , 𝑓𝑜𝑟 𝛾
1 𝐻
𝐻
𝐻𝐼
𝐻𝐼𝐼
4(1 − 𝛾)(1 − 𝑆𝑝 ) (1 + 1 − 𝑆 𝐻𝐼𝐼,𝑇 ) 𝐼𝐼
]
𝑝
𝐼𝐼
(
∈ [0,1)
where 𝑝𝑡𝑡,𝐸𝑈𝑇 ∈ [0,1] is the EUT based testing threshold.
3
Treatment threshold
We set 𝑉(𝑅𝑥) = 𝑉(𝑇) which results in:
𝛾
𝛾
(𝑈𝐼,2 − 𝑈𝐼,4 ) + (1 − 𝛾)[𝑝𝑈𝐼𝐼,1 + (1 − 𝑝)𝑈𝐼𝐼,2 ] = (𝑈𝐼,2 − 𝑈𝐼,4 + 𝑈𝐼,3 − 𝑈𝐼,1 )
2
4
+(1 − 𝛾)[𝑝𝑆𝑈𝐼𝐼,1 + (1 − 𝑝)(1 − 𝑆𝑝 )𝑈𝐼𝐼,2 + 𝑝(1 − 𝑆)𝑈𝐼𝐼,3 + (1 − 𝑝)𝑆𝑝 𝑈𝐼𝐼,4 ]
− (𝛾𝐻𝐼,𝑇 + (1 − 𝛾)𝐻𝐼𝐼,𝑇 )
𝛾
(𝐵 − 𝐻𝐼 ) + (𝛾𝐻𝐼,𝑇 + (1 − 𝛾)𝐻𝐼𝐼,𝑇 )
4 𝐼
= (1 − 𝛾)[−𝑝(1 − 𝑆)𝐵𝐼𝐼 − (1 − 𝑝)𝑆𝑝 𝑈𝐼𝐼,2 + (1 − 𝑝)𝑆𝑝 𝑈𝐼𝐼,4 ]
𝑝[(1 − 𝑆)𝐵𝐼𝐼 + 𝑆𝑝 𝐻𝐼𝐼 ] = 𝑆𝑝 𝐻𝐼𝐼 −
𝑝𝑟𝑥 =
𝑆𝑝 𝐻𝐼𝐼 − 𝐻𝐼𝐼,𝑇
𝑆𝑝 𝐻𝐷
((1 − 𝑆)𝐵𝐼𝐼 + 𝑆𝑝 𝐻𝐼𝐼 )
𝑆𝑝 𝐻𝐼𝐼
𝑝𝑟𝑥
+
(𝛾𝐻𝐼𝑇 + (1 − 𝛾)𝐻𝐼𝐼,𝑇 )
𝛾
(𝐵𝐼 − 𝐻𝐼 ) −
(1 − 𝛾)
4(1 − 𝛾)
(𝐻𝐼 − 𝐵𝐼 )
𝛾
𝐻𝐼,𝑇
(
−4
)
(1 − 𝑆)𝐵𝐼𝐼 + 𝑆𝑝 𝐻𝐼𝐼
𝑆𝑝 𝐻𝐼𝐼
𝑆𝑝 𝐻𝐼𝐼
4(1 − 𝛾) (
)
𝑆𝑝 𝐻𝐼𝐼
1𝐻
1 − 𝑆 𝐻𝐼𝐼,𝑇
𝛾
𝐻𝐼
𝐵𝐼
𝐻𝐼,𝑇
𝑝
𝐼𝐼
=
1+
( (1 − ) − 4
)
1−𝑆𝐵
1𝐻
𝐻
𝐻𝐼
𝐻𝐼𝐼
1 + 𝑆 𝐻𝐼𝐼
4(1 − 𝛾)𝑆𝑝 (1 − 𝑆 𝐻𝐼𝐼,𝑇 ) 𝐼𝐼
𝑝
𝐼𝐼 [
]
𝑝
𝐼𝐼
𝑝𝑟𝑥 = 𝑝𝑟𝑥,𝐸𝑈𝑇 1 +
[
𝛾
𝐻𝐼
𝐵𝐼
𝐻𝐼,𝑇
( (1 − ) − 4
)
1 𝐻𝐼𝐼,𝑇 𝐻𝐼𝐼
𝐻𝐼
𝐻𝐼𝐼
4(1 − 𝛾)𝑆𝑝 (1 − 𝑆 𝐻 )
]
𝑝
𝐼𝐼
To constrain the values of 𝑝𝑟𝑥 ∈ [0,1], we modify the 𝑝𝑟𝑥 formula such as:
𝑝𝑟𝑥 = min {𝑝𝑟𝑥,𝐸𝑈𝑇 1 +
[
𝛾
𝐻𝐼
𝐵𝐼
𝐻𝐼,𝑇
(1 − ) − 4
) , 1} , 𝑓𝑜𝑟 𝛾
1 𝐻𝐼𝐼,𝑇 𝐻𝐼𝐼
𝐻𝐼
𝐻𝐼𝐼
4(1 − 𝛾)𝑆𝑝 (1 − 𝑆 𝐻 )
]
𝑝
𝐼𝐼
(
∈ [0,1)
where 𝑝𝑟𝑥,𝐸𝑈𝑇 ∈ [0,1] is the EUT based testing threshold.
Special Case: 𝛾 = 1
When 𝛾 = 1, equations 1, 2, and 3 of the main manuscript are undefined. However, we
can still identify the optimal decision by re-deriving the equations for 𝑝𝑡 , 𝑝𝑡𝑡 , and 𝑝𝑟𝑥
4
using 𝛾 = 1 in each strategy valuation function. The strategy with greatest valuation
(see Fig 3) corresponds to the optimal decision. The valuations for each strategy are:
1
1
𝑉(𝑁𝑜𝑅𝑥) = 2 (𝑈𝐼,3 − 𝑈𝐼,1 ) = − 2 𝐵𝐼
(A1)
𝛾
1
𝑉(𝑅𝑥) = 2 (𝑈𝐼,2 − 𝑈𝐼,4 ) + (1 − 𝛾)[𝑝𝑈𝐼𝐼,1 + (1 − 𝑝)𝑈𝐼𝐼,2 ] = − 2 𝐻𝐼
1
1
1
𝑉(𝑇) = 4 (𝑈𝐼,2 − 𝑈𝐼,4 + 𝑈𝐼,3 − 𝑈𝐼,1 ) − 𝐻𝐼,𝑇 = − 4 𝐻𝐼 − 4 𝐵𝐼 − 𝐻𝐼,𝑇
(A2)
(A3)
1. For equation 1 of the main manuscript, which is used to decide between treating
and not treating, we choose to treat when (𝐴2) ≥ (𝐴1) → 𝑉(𝑅𝑥) ≥ 𝑉(𝑁𝑜𝑅𝑥) →
𝐵𝐼 ≥ 𝐻𝐼 otherwise we choose not to treat .
2. Similarly, for equation 2 of the main manuscript, which is used to decide between
testing and no treatment, we choose no treatment when (𝐴1) ≥ (𝐴3) →
𝑉(𝑁𝑜𝑅𝑥) ≥ 𝑉(𝑇) → 𝐵𝐼 ≤ 𝐻𝐼 + 4𝐻𝐼,𝑇 . Conversely, we choose testing when 𝐵𝐼 >
𝐻𝐼 + 4𝐻𝐼,𝑇 .
3. Finally, for equation 3 of the main manuscript, which is used to decide between
treating and testing, we choose treatment when (𝐴2) ≥ (𝐴3) → 𝑉(𝑅𝑥) ≥ 𝑉(𝑇) →
𝐵𝐼 ≥ 𝐻𝐼 − 4𝐻𝐼,𝑇 and we choose testing otherwise.
Since 𝐻𝐼 − 4𝐻𝐼,𝑇 ≤ 𝐻𝐼 + 4𝐻𝐼,𝑇 , items 2 and 3 above can be combined as follows:

From item 2, we choose testing when 𝐵𝐼 > 𝐻𝐼 + 4𝐻𝐼,𝑇 . However, if 𝐵𝐼 > 𝐻𝐼 + 4𝐻𝐼,𝑇
then 𝐵𝐼 > 𝐻𝐼 − 4𝐻𝐼,𝑇 . Therefore, from item 3 above we can skip testing and
choose treatment instead.

From item 3, we choose testing when 𝐵𝐼 < 𝐻𝐼 − 4𝐻𝐼,𝑇 . However, if 𝐵𝐼 < 𝐻𝐼 − 4𝐻𝐼,𝑇
then 𝐵𝐼 < 𝐻𝐼 + 4𝐻𝐼,𝑇 . Therefore, from item 2 above, we can skip testing and
choose not to treat instead.
Note that whenever 𝐻𝐼 − 4𝐻𝐼,𝑇 < 𝐵𝐼 < 𝐻𝐼 + 4𝐻𝐼,𝑇 , we are indifferent between picking
treatment or no treatment, but since we don’t want to test, we can simply assume 𝐻𝐼,𝑇 =
0 and base our decision on the comparison of 𝐵𝐼 and 𝐻𝐼 .
To summarize, when 𝛾 = 1, we choose treatment when 𝐵𝐼 ≥ 𝐻𝐼 − 4𝐻𝐼,𝑇 and choose no
treatment if 𝐵𝐼 ≤ 𝐻𝐼 + 4𝐻𝐼,𝑇 .
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