Estimating a probability given other probabilities from a prior

Question

I have a bunch of data coming in (calls to an automated callcenter) about whether or not a person buys a particular product, 1 for buy, 0 for not buy.

I want to use this data to create an estimated probability that a person will buy a particular product, but the problem is that I may need to do it with relatively little historical data about how many people bought/didn't buy that product.

A friend recommended that with Bayesian probability you can "help" your probability estimate by coming up with a "prior probability distribution", essentially this is information about what you expect to see, prior to taking into account the actual data.

So what I'd like to do is create a method that has something like this signature (Java):

double estimateProbability(double[] priorProbabilities, int buyCount, int noBuyCount);

priorProbabilities is an array of probabilities I've seen for previous products, which this method would use to create a prior distribution for this probability. buyCount and noBuyCount are the actual data specific to this product, from which I want to estimate the probability of the user buying, given the data and the prior. This is returned from the method as a double.

I don't need a mathematically perfect solution, just something that will do better than a uniform or flat prior (ie. probability = buyCount / (buyCount+noBuyCount)). Since I'm far more familiar with source code than mathematical notation, I'd appreciate it if people could use code in their explanation.

Answer 1

Here's the Bayesian computation and one example/test:

def estimateProbability(priorProbs, buyCount, noBuyCount):
  # first, estimate the prob that the actual buy/nobuy counts would be observed
  # given each of the priors (times a constant that's the same in each case and
  # not worth the effort of computing;-)`
  condProbs = [p**buyCount * (1.0-p)**noBuyCount for p in priorProbs]
  # the normalization factor for the above-mentioned neglected constant
  # can most easily be computed just once
  normalize = 1.0 / sum(condProbs)
  # so here's the probability for each of the prior (starting from a uniform
  # metaprior)
  priorMeta = [normalize * cp for cp in condProbs]
  # so the result is the sum of prior probs weighed by prior metaprobs
  return sum(pm * pp for pm, pp in zip(priorMeta, priorProbs))

def example(numProspects=4):
  # the a priori prob of buying was either 0.3 or 0.7, how does it change
  # depending on how 4 prospects bought or didn't?
  for bought in range(0, numProspects+1):
    result = estimateProbability([0.3, 0.7], bought, numProspects-bought)
    print 'b=%d, p=%.2f' % (bought, result)

example()

output is:

b=0, p=0.31
b=1, p=0.36
b=2, p=0.50
b=3, p=0.64
b=4, p=0.69

which agrees with my by-hand computation for this simple case. Note that the probability of buying, by definition, will always be between the lowest and the highest among the set of priori probabilities; if that's not what you want you might want to introduce a little fudge by introducing two "pseudo-products", one that nobody will ever buy (p=0.0), one that anybody will always buy (p=1.0) -- this gives more weight to actual observations, scarce as they may be, and less to statistics about past products. If we do that here, we get:

b=0, p=0.06
b=1, p=0.36
b=2, p=0.50
b=3, p=0.64
b=4, p=0.94

Intermediate levels of fudging (to account for the unlikely but not impossible chance that this new product may be worse than any one ever previously sold, or better than any of them) can easily be envisioned (give lower weight to the artificial 0.0 and 1.0 probabilities, by adding a vector priorWeights to estimateProbability's arguments).

This kind of thing is a substantial part of what I do all day, now that I work developing applications in Business Intelligence, but I just can't get enough of it...!-)

Answer 2

A really simple way of doing this without any difficult math is to increase buyCount and noBuyCount artificially by adding virtual customers that either bought or didn't buy the product. You can tune how much you believe in each particular prior probability in terms of how many virtual customers you think it is worth.

In pseudocode:

def estimateProbability(priorProbs, buyCount, noBuyCount, faithInPrior=None):
    if faithInPrior is None: faithInPrior = [10 for x in buyCount]
    adjustedBuyCount = [b + p*f for b,p,f in 
                                zip(buyCount, priorProbs, faithInPrior]
    adjustedNoBuyCount = [n + (1-p)*f for n,p,f in 
                                zip(noBuyCount, priorProbs, faithInPrior]
    return [b/(b+n) for b,n in zip(adjustedBuyCount, adjustedNoBuyCount]