The odds form of Bayes' rule

We've seen that the likelihood ratio compares how probable the evidence $E$ is under $H$ and under $H^c$. What we usually want is the conditional probability the other way round: $P(H \mid E)$, the posterior probability of $H$ once $E$ is observed.

Bayes' rule gives $P(H \mid E)$, but it needs the denominator $P(E)$, which usually takes the law of total probability to find. Written in terms of odds, Bayes' rule becomes simpler: $P(E)$ cancels, and the evidence enters through a single multiplication by the likelihood ratio.

To write Bayes' rule in terms of odds, we need the odds of $H$ before and after $E$ is observed.

Bayes' rule gives both posterior probabilities with the same denominator, $P(E)$:

Dividing the first by the second cancels $P(E)$:

In words, the posterior odds are the prior odds times the likelihood ratio. Before $E$ is observed, the odds of $H$ are the prior odds. Observing $E$ updates them by multiplying by the likelihood ratio: evidence that favours $H$ has a ratio above $1$ and raises the odds, and evidence against $H$ has a ratio below $1$ and lowers them.

A machine part is faulty with probability $0.2$. A quality check flags $0.9$ of faulty parts and $0.15$ of parts that work. Find the odds that a flagged part is faulty.

Solution

We name the hypothesis and the evidence:

• $H$: the part is faulty.
• $E$: the check flags the part.

The prior is $P(H) = 0.2$, so $P(H^c) = 0.8$ and the prior odds are

Each figure is a likelihood, $P(E \mid H) = 0.9$ and $P(E \mid H^c) = 0.15$, so the likelihood ratio is

Multiplying the prior odds by it gives the posterior odds:

The odds that a flagged part is faulty are $3$ to $2$.

The odds form of Bayes' rule gives the posterior odds of $H$, but often what we want is the posterior probability $P(H \mid E)$.

Recall that when the odds of an event are a single number $o$, its probability is $\dfrac{o}{1 + o}$. So once we have the posterior odds, $o = \text{odds}(H \mid E)$, the posterior probability is

Of a bank's transactions, $0.01$ are fraudulent. An alert is raised $33$ times as often on fraudulent transactions as on genuine ones. Find the probability that an alerted transaction is fraudulent.

Solution

We name the hypothesis and the evidence:

• $H$: the transaction is fraudulent.
• $E$: an alert is raised.

The prior is $P(H) = 0.01$, so $P(H^c) = 0.99$ and the prior odds are

An alert is raised $33$ times as often when $H$ holds as when it does not, so the likelihood ratio is $33$. Multiplying the prior odds by it gives the posterior odds:

We convert the posterior odds $o = \dfrac{1}{3}$ to a probability with $\dfrac{o}{1 + o}$:

An alerted transaction is fraudulent with probability $0.25$.