Two independent events become conditionally dependent given that at least one of them occurs. Symbolically:
- If
and
then
![{\displaystyle P(A\cap B|A\cup B)<P(A|A\cup B)P(B|A\cup B)}](//wikimedia.org/api/rest_v1/media/math/render/svg/954095091de7dfa59f606c9f34b00c3614f888e5)
Proof: Note that
and
which, together with
and
(so
) implies that
![{\displaystyle {\begin{aligned}P(A\cap B|A\cup B)={\frac {P(A\cap B)}{P(A\cup B)}}={\frac {P(A)P(B)}{P(A\cup B)}}<{\frac {P(A)P(B)}{[P(A\cup B)]^{2}}}=P(A|A\cup B)P(B|A\cup B).\end{aligned}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/fb0b65ed62e157f1a5c42ed19367a143d08a33a4)
One can see this in tabular form as follows: the yellow regions are the outcomes where at least one event occurs (and ~A means "not A").
More information A, ~A ...
| A | ~A |
B |
A & B |
~A & B |
~B |
A & ~B |
~A & ~B |
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For instance, if one has a sample of
, and both
and
occur independently half the time (
), one obtains:
More information A, ~A ...
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So in
outcomes, either
or
occurs, of which
have
occurring. By comparing the conditional probability of
to the unconditional probability of
:
![{\displaystyle P(A|A\cup B)=50/75=2/3>P(A)=50/100=1/2}](//wikimedia.org/api/rest_v1/media/math/render/svg/f974917f681c6faeab581184a617d016726d369e)
We see that the probability of
is higher (
) in the subset of outcomes where (
or
) occurs, than in the overall population (
). On the other hand, the probability of
given both
and (
or
) is simply the unconditional probability of
,
, since
is independent of
. In the numerical example, we have conditioned on being in the top row:
More information A, ~A ...
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Here the probability of
is
.
Berkson's paradox arises because the conditional probability of
given
within the three-cell subset equals the conditional probability in the overall population, but the unconditional probability within the subset is inflated relative to the unconditional probability in the overall population, hence, within the subset, the presence of
decreases the conditional probability of
(back to its overall unconditional probability):
![{\displaystyle P(A|B,A\cup B)=P(A|B)=P(A)}](//wikimedia.org/api/rest_v1/media/math/render/svg/e62af89037d80e9a390687d7573337a280decf5c)
![{\displaystyle P(A|A\cup B)>P(A)}](//wikimedia.org/api/rest_v1/media/math/render/svg/618d4bdd0e4906f14b14a4ffe4d9b1721178854e)
Because the effect of conditioning on
derives from the relative size of
and
the effect is particularly large when
is rare (
) but very strongly correlated to
(
). For example, consider the case below where N is very large:
More information A, ~A ...
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For the case without conditioning on
we have
![{\displaystyle P(A)=1/(N+1)}](//wikimedia.org/api/rest_v1/media/math/render/svg/3e45fc1842eb92594e884de5843dc905dfc7922f)
![{\displaystyle P(A|B)=1}](//wikimedia.org/api/rest_v1/media/math/render/svg/44773125188949241ab05f6976148c13c62ddb65)
So A occurs rarely, unless B is present, when A occurs always. Thus B is dramatically increasing the likelihood of A.
For the case with conditioning on
we have
![{\displaystyle P(A|A\cup B)=1}](//wikimedia.org/api/rest_v1/media/math/render/svg/71cefa3c2fd0af89fb3aedec940fa8400b4aeb1d)
![{\displaystyle P(A|B,A\cup B)=P(A|B)=1}](//wikimedia.org/api/rest_v1/media/math/render/svg/ad8fb4de1e3fdeb29a6e5361eaa1ae87ce2a8346)
Now A occurs always, whether B is present or not. So B has no impact on the likelihood of A. Thus we
see that for highly correlated data a huge positive correlation of B on A can be effectively removed when one conditions on
.