5
$\begingroup$

Say we want to build a classifier for a binary classification problem using a discriminative method (e.g. SVM) and be able to impose a prior on the classes.

For example, let's assume that we want to use the prior $\text{Beta}(10,50)$ on the positive class.

How can I estimate the posterior probability of classification resulting from combining the output of my discriminative predictor with the above class prior?

$\endgroup$
1
$\begingroup$

I'm going to assume that you've trained your classifier using equal numbers of samples where $y=0$ and $y=1$, that is, with an implicit prior that $\frac{P_{Old}(y=1)}{P_{Old}(y=0)} = \frac{1}{1}$.

I'll also assume that your classifier makes probabilistic predictions about $P_{Old}(y=1|x, \theta)$, where $x$ is the to-be-predicted data and $\theta$ are the trained parameters/support vectors. It's useful to represent these predictions as odds ratios, $\frac{P_{Old}(y=1|x, \theta)}{P_{Old}(y=0|x, \theta)}$.

To obtain the posterior odds given your new prior, we need to find the likelihood ratio, $\frac{P(x|y=1, \theta)}{P(x|y=0, \theta)}$. Since your old prior odds were $\frac{1}{1}$, the likelihood ratio is the same as your old prediction odds ratio:

$$ \tag{1} \frac{P(x|y=1, \theta)}{P(x|y=0, \theta)} = \frac{1}{1} \times \frac{P_{Old}(y=1|x, \theta)}{P_{Old}(y=0|x, \theta)}; $$

Now, imagine first that your new prior odds are just a point estimate, $\frac{P_{New}(y=1)}{P_{New}(y=0)} = \frac{\pi_1}{\pi_0}$. Your new posterior odds are just the new prior odds multiplied by the likelihood ratio, which is also the prediction from your SVM:

$$ \tag{2} \frac{P_{New}(y=1|x, \theta)}{P_{New}(y=0|x, \theta)} = \frac{\pi_1}{\pi_0} \times \frac{P_{Old}(y=1|x, \theta)}{P_{Old}(y=0|x, \theta)} $$

You instead have a distribution over prior probabilities, $P(y=1) \sim \text{Beta}(\alpha, \beta)$. There are a few ways to obtain the corresponding posterior, but the easiest is to sample class probabilities from the prior, convert to odds, and multiply by the likelihood to obtain samples from the posterior. Alternatively, you can calculate the density over a grid of prior probabilities, and transform appropriately to find the corresponding posterior probabilities. The code below demonstrates both approaches.

library(magrittr)
odds2prob = function(o) o / (1 + o)
prob2odds = function(p) p / (1 - p)

old.prior.odds = 1/1
svm.posterior.prob = .8 # P(y=1)
svm.posterior.odds = prob2odds(svm.posterior.prob) # = 4/1
likelihood.ratio = svm.posterior.odds / old.prior.odds # Also = 4/1

# New odds point estimate
new.prior.odds = 1/3
new.posterior.odds = new.prior.odds * likelihood.ratio # 4/1 * 1/3 = 4/3
new.posterior.odds

# Samples from posterior
prior.prob.samps = rbeta(1000, 10, 20)
prior.odds.samps = prob2odds(prior.prob.samps)
new.posterior.odds.samps = prior.odds.samps * likelihood.ratio
# Take the mean of the log odds to Normalise the distribution
new.posterior.prediction = new.posterior.odds.samps %>% log() %>% mean() %>% exp() %>% odds2prob()
new.posterior.prediction # ~0.66

# Calculate over grid
probs = seq(0.001, .999, .01)
density = dbeta(probs, 10, 20)

odds = prob2odds(probs)
plot(probs, density, main='Prior density', xlab = 'P(y=1)')
plot(odds2prob(odds * likelihood.ratio), density, 
     main='Posterior density', xlab = 'P(y=1)')

enter image description here enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks Eoin, do you mind further clarifying how you establish the equality in Eq. 1 from first principles? i.e. why is LHS = RHS in that equation? $\endgroup$ – Amelio Vazquez-Reina Jul 24 at 20:54

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.