A
case
probability
of
misleading
Today we are going to see another of those examples where intuition
about the value of certain probabilities plays tricks on us. And, for that,
we will use nothing less than Bayes’ theorem, playing a little with
conditioned probabilities. Let’s see step by step how it works.
What is the probability of two events occurring? The probability of an
event A occurring is P(A) and that of B, P(B). Well, the probability of the
two occurring is P(A∩B), which, if the two events are independent, is equal
to P(A) x P(B).
Imagine that we have a die with six faces. If we throw it once, the
probability of taking out, for example, a five is 1/6 (one result among the
six possible). The probability to draw a four is also 1/6. What will be the
probability of getting a four, once in the first roll we get a five? Since
the two runs are independent, the probability of the combination five
followed by four will be 1/6 x 1/6 = 1/36.
Now let’s think of another example. Suppose that in a group of 10 people
there are four doctors, two of whom are surgeons. If we take one at random,
the probability of being a doctor is 4/10 = 0.4 and that of a surgeon is
2/10 = 0.2. But if we get one and know that he is a doctor, the probability
that he is a surgeon will no longer be 0.2, because the two events, being a
doctor and a surgeon, are not independent. If you are a doctor, the
probability that you are a surgeon will be 0.5 (half the doctors in our
group are surgeons).
When two events are dependent, the probability of occurrence of the two
will be the probability of occurrence of the first, once the second occurs,
by the probability of occurrence of the second. So the P(surgeon) =
P(surgeon|doctor) x P(doctor). We can generalize the expression as follows:
P(A∩B) = P(A|B) x P(B), and changing the order of the components of the
expression, we obtain the so-called Bayes rule, as follows:
P(A|B) = P(A∩B) / P(B).
The P(A∩B) will be the probability of B, once A is produced, by the
probability of A = P(B|A) x P(A). On the other hand, the probability of B
will be equal to the sum of the probability of occurrence B once A is
produced plus the probability of occurring B without occurring A, which put
in mathematical form is of the following form:
P(B|A) x P(A) + P(B|A c ) x P(A c ), being P(A c ) the probability of not
occurring A.
If we substitute the initial rule for its developed values, we obtain
the best known expression of the Bayes theorem:
Let’s see how the Bayes theorem is
applied with a practical example. Consider the case of acute
fildulastrosis, a serious disease whose prevalence in the population is,
fortunately, quite low, one per 1000 inhabitants. Then, the P(F) = 0.001.
Luckily we have a good diagnostic test, with a sensitivity of 98% and a
specificity of 95%. Suppose now that I take the test and it gives me a
positive result. Do I have to scare myself a lot? What is the probability
that I actually have the disease? Do you think it will be high or low?
Let’s see.
A sensitivity of 98% means that the probability of giving positive when
having the disease is 0.98. Mathematically, P(POS|F) = 0,98. On the other
hand, a specificity of 95% means that the probability of a negative result
being healthy is 0.95. That is, P(NEG|Fc) = 0.95. But what we want to know
is neither of these two things, but we really look for the probability of
being sick once we test positive, that is, P (F|POS).
To calculate it, we have only to apply the theorem of Bayes:
Then we replace the symbols
with their values and solve the equation:
So
we
see
that,
in
principle, I do not have to scare a lot when the test gives me a positive
result, since the probability of being ill is only 2%. As you see, much
lower than intuition would tell us with such a high sensitivity and
specificity. Why is this happening? Very simple, because the prevalence of
the disease is very low. We are going to repeat the experiment assuming now
that the prevalence is 10% (0,1):
As you see, in this case the
probability of being ill if I give positive rises to 68%. This probability
is known as positive predictive value which, as we can see, can vary
greatly depending on the frequency of the effect we are studying.
And here we leave it for today. Before closing, let me warn you not to
seek what the fildulastrosis is. I would be very surprised if anyone found
it in a medical book. Also, be careful not to confuse P (POS|F) with P
(F|POS), since you would make a mistake called reverse fallacy or fallacy
of transposition of conditionals, which is a serious error.
We have seen how the calculation of probabilities gets somewhat
complicated when the events are not independent. We have also learned how
unreliable predictive values are when the prevalence of the disease
changes. That is why the likelihood ratios were invented, which do not
depend so much on the prevalence of the disease that is diagnosed and allow
a better overall assessment of the power of the diagnostic test. But that
is another story…