Conditional probability and Bayes' rule

A trading signal is evidence. The question is never “is the signal on?” but “given that the signal is on, how likely is the outcome I care about?” That is a conditional probability.

Conditional probability

The probability of $A$ given that $B$ has happened is

P(A \mid B) = \frac{P(A \cap B)}{P(B)}

You restrict the universe to the cases where $B$ is true, and ask what fraction of those also have $A$ .

Bayes’ rule

Rearranging the definition both ways gives Bayes’ rule:

P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B)}

with the denominator usually expanded by the law of total probability:

P(B) = P(B \mid A)P(A) + P(B \mid A^c)P(A^c)

The pieces have names. $P(A)$ is the prior (what you believed before), $P(B \mid A)$ is the likelihood (how expected the evidence is if $A$ is true), and $P(A \mid B)$ is the posterior.

Worked example: a “95% accurate” signal

A crash-warning model flags 95% of real crashes, and falsely flags 10% of normal weeks. Crashes happen in 2% of weeks. The model is flashing. How worried should you be?

P(\text{crash} \mid \text{flag}) = \frac{0.95 \times 0.02}{0.95 \times 0.02 + 0.10 \times 0.98} = \frac{0.019}{0.117} \approx 16\%

The “95% accurate” model is wrong about five times in six when it fires, because the thing it detects is rare. This is base-rate neglect, and it explains most disappointing signals.

Thinking in odds

Bayes is cleaner in odds form. Writing $O(A) = P(A)/P(A^c)$ :

O(A \mid B) = O(A) \times \frac{P(B \mid A)}{P(B \mid A^c)}

Posterior odds equal prior odds times the likelihood ratio. In the example the prior odds are $2{:}98$ , the likelihood ratio is $9.5$ , so the posterior odds are $19{:}98$ , about 16%. Each new independent piece of evidence simply multiplies in another likelihood ratio, which is why combining several weak, independent signals can be powerful.

Independence

$A$ and $B$ are independent when $P(A \mid B) = P(A)$ , equivalently $P(A \cap B) = P(A)P(B)$ . Markets break independence constantly: large moves cluster, assets that look unrelated move together in a crisis. Treat independence as an assumption to be checked, never a default.

Common mistake

Confusing $P(B \mid A)$ with $P(A \mid B)$ . “Most big winners had high momentum” is $P(\text{momentum} \mid \text{winner})$ . What you trade on is $P(\text{winner} \mid \text{momentum})$ , which can be far smaller if most high-momentum stocks don’t become big winners.

Interview classics

Two children. A family has two children and at least one is a boy. The probability both are boys is $1/3$ , not $1/2$ : of the equally likely cases BB, BG, GB, only one of the three remaining is BB.
Monty Hall. You pick a door; the host, who knows where the car is, opens a different door showing a goat. Switching wins with probability $2/3$ , because your first pick was right only $1/3$ of the time and the host’s action never changes that.
The biased coin. One of 100 coins is double-headed. You draw one at random and flip ten heads. The posterior that it’s the double-headed coin is $\frac{1/100}{1/100 + (99/100)(1/2)^{10}} \approx 91\%$ .

Practise more in the question bank.