17 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Differential diagnosis | 3/6 | https://en.wikipedia.org/wiki/Differential_diagnosis | reference | science, encyclopedia | 2026-05-05T07:27:56.437702+00:00 | kb-cron |
Pr
(
PH in population
)
=
0.5
years
⋅
1
4000 per year
=
1
8000
{\displaystyle \Pr({\text{PH in population}})=0.5{\text{ years}}\cdot {\frac {1}{\text{4000 per year}}}={\frac {1}{8000}}}
With the relative risk conferred from the family history, the probability that primary hyperparathyroidism (PH) would have occurred in the first place in the individual given from the currently available information becomes:
Pr
(
PH WHOIFPI
)
≈
R
R
P
H
⋅
Pr
(
PH in population
)
=
10
⋅
1
8000
=
1
800
=
0.00125
{\displaystyle \Pr({\text{PH WHOIFPI}})\approx RR_{PH}\cdot \Pr({\text{PH in population}})=10\cdot {\frac {1}{8000}}={\frac {1}{800}}=0.00125}
Primary hyperparathyroidism can be assumed to cause hypercalcemia essentially 100% of the time (rPH → hypercalcemia = 1), so this independently calculated probability of primary hyperparathyroidism (PH) can be assumed to be the same as the probability of being a cause of the presentation:
Pr
(
Hypercalcemia WHOIFPI by PH
)
=
Pr
(
PH WHOIFPI
)
⋅
r
PH
→
hypercalcemia
=
0.00125
⋅
1
=
0.00125
{\displaystyle {\begin{aligned}\Pr({\text{Hypercalcemia WHOIFPI by PH}})&=\Pr({\text{PH WHOIFPI}})\cdot r_{{\text{PH}}\rightarrow {\text{hypercalcemia}}}\\&=0.00125\cdot 1=0.00125\end{aligned}}}
For cancer, the same time-at-risk is assumed for simplicity, and let's say that the incidence of cancer in the area is estimated at 1 in 250 per year, giving a population probability of cancer of:
Pr
(
cancer in population
)
=
0.5
years
⋅
1
250 per year
=
1
500
{\displaystyle \Pr({\text{cancer in population}})=0.5{\text{ years}}\cdot {\frac {1}{\text{250 per year}}}={\frac {1}{500}}}
For simplicity, let's say that any association between a family history of primary hyperparathyroidism and risk of cancer is ignored, so the relative risk for the individual to have contracted cancer in the first place is similar to that of the population (RRcancer = 1):
Pr
(
cancer WHOIFPI
)
≈
R
R
cancer
⋅
Pr
(
cancer in population
)
=
1
⋅
1
500
=
1
500
=
0.002.
{\displaystyle \Pr({\text{cancer WHOIFPI}})\approx RR_{\text{cancer}}\cdot \Pr({\text{cancer in population}})=1\cdot {\frac {1}{500}}={\frac {1}{500}}=0.002.}
However, hypercalcemia only occurs in, very approximately, 10% of cancers, (rcancer → hypercalcemia = 0.1), so:
Pr
(
Hypercalcemia WHOIFPI by cancer
)
=
Pr
(
cancer WHOIFPI
)
⋅
r
cancer
→
hypercalcemia
=
0.002
⋅
0.1
=
0.0002.
{\displaystyle {\begin{aligned}&\Pr({\text{Hypercalcemia WHOIFPI by cancer}})\\=&\Pr({\text{cancer WHOIFPI}})\cdot r_{{\text{cancer}}\rightarrow {\text{hypercalcemia}}}\\=&0.002\cdot 0.1=0.0002.\end{aligned}}}
The probabilities that hypercalcemia would have occurred in the first place by other candidate conditions can be calculated in a similar manner. However, for simplicity, let's say that the probability that any of these would have occurred in the first place is calculated at 0.0005 in this example. For the instance of there being no disease, the corresponding probability in the population is complementary to the sum of probabilities for other conditions:
Pr
(
no disease in population
)
=
1
−
Pr
(
PH in population
)
−
Pr
(
cancer in population
)
−
Pr
(
other conditions in population
)
=
0.997.
{\displaystyle {\begin{aligned}\Pr({\text{no disease in population}})&=1-\Pr({\text{PH in population}})-\Pr({\text{cancer in population}})\\&{}\quad -\Pr({\text{other conditions in population}})\\&{}=0.997.\end{aligned}}}
The probability that the individual would be healthy in the first place can be assumed to be the same:
Pr
(
no disease WHOIFPI
)
=
0.997.
{\displaystyle \Pr({\text{no disease WHOIFPI}})=0.997.\,}
The rate at which the case of no abnormal condition still ends up in measurement of serum calcium of being above the standard reference range (thereby classifying as hypercalcemia) is, by the definition of standard reference range, less than 2.5%. However, this probability can be further specified by considering how much the measurement deviates from the mean in the standard reference range. Let's say that the serum calcium measurement was 1.30 mmol/L, which, with a standard reference range established at 1.05 to 1.25 mmol/L, corresponds to a standard score of 3 and a corresponding probability of 0.14% that such degree of hypercalcemia would have occurred in the first place in the case of no abnormality:
r
no disease
→
hypercalcemia
=
0.0014
{\displaystyle r_{{\text{no disease}}\rightarrow {\text{hypercalcemia}}}=0.0014}
Subsequently, the probability that hypercalcemia would have resulted from no disease can be calculated as:
Pr
(
Hypercalcemia WHOIFPI by no disease
)
=
Pr
(
no disease WHOIFPI
)
⋅
r
no disease
→
hypercalcemia
=
0.997
⋅
0.0014
≈
0.0014
{\displaystyle {\begin{aligned}&\Pr({\text{Hypercalcemia WHOIFPI by no disease}})\\=&\Pr({\text{no disease WHOIFPI}})\cdot r_{{\text{no disease}}\rightarrow {\text{hypercalcemia}}}\\=&0.997\cdot 0.0014\approx 0.0014\end{aligned}}}
The probability that hypercalcemia would have occurred in the first place in the individual can thus be calculated as:
Pr
(
hypercalcemia WHOIFPI
)
=
Pr
(
hypercalcemia WHOIFPI by PH
)
+
Pr
(
hypercalcemia WHOIFPI by cancer
)
+
Pr
(
hypercalcemia WHOIFPI by other conditions
)
+
Pr
(
hypercalcemia WHOIFPI by no disease
)
=
0.00125
+
0.0002
+
0.0005
+
0.0014
=
0.00335
{\displaystyle {\begin{aligned}&\Pr({\text{hypercalcemia WHOIFPI}})\\=&\Pr({\text{hypercalcemia WHOIFPI by PH}})+\Pr({\text{hypercalcemia WHOIFPI by cancer}})\\&{}+\Pr({\text{hypercalcemia WHOIFPI by other conditions}})+\Pr({\text{hypercalcemia WHOIFPI by no disease}})\\=&0.00125+0.0002+0.0005+0.0014=0.00335\end{aligned}}}
Subsequently, the probability that hypercalcemia is caused by primary hyperparathyroidism (PH) in the individual can be calculated as:
Pr
(
hypercalcemia is caused by PH in individual
)
=
Pr
(
hypercalcemia WHOIFPI by PH
)
Pr
(
hypercalcemia WHOIFPI
)
=
0.00125
0.00335
=
0.373
=
37.3
%
{\displaystyle {\begin{aligned}&\Pr({\text{hypercalcemia is caused by PH in individual}})\\=&{\frac {\Pr({\text{hypercalcemia WHOIFPI by PH}})}{\Pr({\text{hypercalcemia WHOIFPI}})}}\\=&{\frac {0.00125}{0.00335}}=0.373=37.3\%\end{aligned}}}
Similarly, the probability that hypercalcemia is caused by cancer in the individual can be calculated as:
Pr
(
hypercalcemia is caused by cancer in individual
)
=
Pr
(
hypercalcemia WHOIFPI by cancer
)
Pr
(
hypercalcemia WHOIFPI
)
=
0.0002
0.00335
=
0.060
=
6.0
%
,
{\displaystyle {\begin{aligned}&\Pr({\text{hypercalcemia is caused by cancer in individual}})\\=&{\frac {\Pr({\text{hypercalcemia WHOIFPI by cancer}})}{\Pr({\text{hypercalcemia WHOIFPI}})}}\\=&{\frac {0.0002}{0.00335}}=0.060=6.0\%,\end{aligned}}}
and for other candidate conditions:
Pr
(
hypercalcemia is caused by other conditions in individual
)
=
Pr
(
hypercalcemia WHOIFPI by other conditions
)
Pr
(
hypercalcemia WHOIFPI
)
=
0.0005
0.00335
=
0.149
=
14.9
%
,
{\displaystyle {\begin{aligned}&\Pr({\text{hypercalcemia is caused by other conditions in individual}})\\=&{\frac {\Pr({\text{hypercalcemia WHOIFPI by other conditions}})}{\Pr({\text{hypercalcemia WHOIFPI}})}}\\=&{\frac {0.0005}{0.00335}}=0.149=14.9\%,\end{aligned}}}
and the probability that there actually is no disease: