0
$\begingroup$


I have been wondering about an issue connected with prevalence. And came up to some conclusions which I would like to verify. It's not complicated. Let's assume we have two prevalences: (A) percentage of population with cancer - 10% and (B) population of smokers - 20%. If there would be no correlation: C = 0, the populatin smokers with cancer (D) would be D = AB = 2%. If The correlation would be C = 100% then everyone with cancer would be a smoker - D = A = 10%. So the correlation proportion wolud be:
C         D
0%      AB
50%    ((A or B) - AB)/2
F%     ((A or B) - AB)*F/100
100%  if A < B then A, if A > B then B

Is this right?
Thank you in adv,
Adi.

$\endgroup$
0
$\begingroup$

Might be easier to think in terms of conditional probability. I'll rephrase your question...

Let $A$ = cancer, and $B$ = smoke

$C$ becomes $P(A|B) - P(A|\neg B)$ and $D$ becomes $P(A \land B)$.

Using Bayes' theorem to expand $C$ and realizing that $D = P(A|B) \cdot P(B)$, you'll get that

$$D = C \cdot P(B) \cdot (1 - P(B)) + P(A)P(B)$$

I think this is more or less what you have... except for the last case since if $C = 1$ then you have cancer iff you smoke, so $P(A) = P(B)$ and $D = P(A) = P(B)$.

Quick derivation scratchbook

P(X|Y) = P(Y|X)P(X) / P(Y)
P(X|\neg Y) = P(\neg Y|X)P(X) / P(\neg Y) = (1 - P(Y|X))P(X) / (1 - P(Y))

C = P(Y|X)P(X) / P(Y)  -  (1 - P(Y|X))P(X) / (1 - P(Y))
= (1 - P(Y)) P(Y|X) P(X)  -  (P(Y)) (1 - P(Y|X)) P(X)  /  (P(Y) (1 - P(Y))
= P(X)P(Y|X) - P(X)P(Y)P(Y|X) - P(X)P(Y) + P(X)P(Y)P(Y|X)  /  P(Y) (1 - P(Y))
= P(X)P(Y|X) - P(X)P(Y)  /  P(Y) (1 - P(Y))

D = P(X|Y) P(Y) = P(Y|X) P(X) / P(Y) * P(Y) = P(Y|X)P(X)

so C = D - P(X)P(Y)   /  P(Y)(1 - P(Y))
D = C * P(Y) * (1 - P(Y)) + P(X)P(Y)
$\endgroup$
  • $\begingroup$ Hi, I really appreciate your help, but this solution is too complicated for me. I heard about conditional probability and planning to read about it but for now I'll stick to my method... $\endgroup$ – Ady Sep 2 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.