## Wednesday, May 26, 2010

### Combining Empirical Hazards by the Naïve Bayesian Method

Occasionally one has an idea that seems so obvious and right that it must surely be standard practice and have a well-known name. A few months ago, I had such an idea while sitting in a client’s office in Ottawa. Last week, I wanted to include the idea in a proposal, so I tried to look it up and couldn’t find a reference to it anywhere. Before letting my ego run away with itself and calling it the naïve Berry model, I figured I would share it with our legions of erudite readers so someone can point me to a reference.

Some Context

My client sells fax-to-email and email-to-fax service on a subscription basis. I had done an analysis to quantify the effect of various factors such as industry code, acquisition channel, and type of phone number (local or long distance) on customer value. Since all customers pay the same monthly fee, the crucial factor is longevity. I had analyzed each covariate separately by calculating cancellation hazard probabilities for each stratum and generating survival curves. The area under the first year of each survival curve is the first year truncated mean tenure. Multiplying the first-year mean tenure by the subscription price yields the average first year revenue for a segment. This let me say how much more valuable a realtor is than a trucker; or a Google adwords referral than an MSN referral.

For many purposes, the dollar value was not even important. We used the probability of surviving one year as a way of scoring particular segments. But how should the individual segment scores be combined to give an individual customer a score based on his being a trucker with an 800 number referred by MSN? Or a tax accountant with a local number referred by Google? The standard empirical hazards approach would be to segment the training data by all levels of all variables before estimating the hazards, but that was not practical since there were so many combinations that many would lack sufficient data to make confident hazard estimates. Luckily, there is a standard model for combining the contributions of several independent pieces of evidence—naïve Bayesian models. An excellent description of the relationship between probability, odds, and likelihood and how to use them to implement naïve Bayesian models, can be found in Chapter 10 of Gordon Linoff’s Data Analysis Using SQL and Excel.

Here are the relevant correspondences:

odds = p/(1-p)
p = 1 - (1/(1+odds))
likelihood = (odds|evidence)/overall odds

Statisticians switch from one representation to another as convenient. A familiar example is logistic regression. Since linear regression is inappropriate for modeling probabilities that range only from 0 to 1, they convert the probabilities to log(odds) that vary from negative infinity to positive infinity. Expressing the log odds as a linear regression equation and solving for p, yields the logistic function.

Naïve Bayesian Models

The Naïve Bayesian model says that the odds of surviving one year given the evidence is the overall odds times the product of the likelihoods for each piece of evidence. For concreteness, let’s calculate a score for a general contractor (industry code 1521) with a local number who was referred by a banner ad.
The probability of surviving one year is 54%. Overall survival odds are therefore 0.54/(1-0.54) or 1.17.
One-year survival for industry code 1521 is 74%, considerably better than overall survival. The survival likelihood is defined as the survival odds, 0.74/(1-0.74) divided by the overall survival odds of 1.17. This works out to 2.43.

One-year survival for local phone numbers is 37%, considerably worse than overall survival. Local phone numbers have one-year survival odds of 0.59 and likelihood of 0.50.

Subscribers acquired through banner ads have one-year survival of 0.52, about the same as overall survival. This corresponds to odds of 1.09 and likelihood of 0.91.

Plugging these values into the naïve Bayesian model formula, we estimate one-year survival odds for this customer as 1.17*2.43*0.50*0.91=1.29. Solving 1.29=p/(p-1) for p yields a one-year survival estimate of 56%, a little bit better than overall survival. The positive evidence from the industry code slightly outweighs the negative evidence from the phone number type.

This example does not illustrate another great feature of naïve Bayesian models. If some evidence is missing—if the subscriber works in an industry for which we have no survival curve, for example—you can simply leave out the industry likelihood term.

The Idea

If we are happy to use the naïve Bayesian model to estimate the probability of a subscriber lasting one year, why not do the same for daily hazard probabilities? This is something I’ve been wanting to do since the first time I ever used the empirical hazard estimation method. That first project was for a wireless phone company. There was plenty of data to calculate hazards stratified by market or rate plan or handset type or credit class or acquisition channel or age group or just about any other time-0 covariate of interest. But there wasn’t enough data to estimate hazards for every combination of the above. I knew about naïve Bayesian models back then; I’d used the Evidence Model in SGI’s Mineset many times. But I never made the connection—it’s hard to combine probabilities, but easy to combine likelihoods. There you have it: Freedom from the curse of dimensionality via the naïve assumption of independence. Estimate hazards for as many levels of as many covariates as you please and then combine them with the naïve Bayesian model. I tried it, and the results were pleasing.

An Example

This example uses data from a mobile phone company. The dataset is available on our web site. There are three rate plans, Top, Middle, and Bottom. There are three markets, Gotham, Metropolis, and Smallville. There are four acquisition channels, Dealer, Store, Chain, and Mail. There is plenty of data to make highly confident hazard estimates for any of the above, but some combinations, such as Smallville-Mail-Top are fairly rare. For many tenures, no one with this combination cancels so there are long stretches of 0 hazard punctuated by spikes where one or two customers leave.

Here are the Smallville-Mail-Top hazard by the Naïve Berry method:

Isn’t that prettier? I think it makes for a prettier survival curve as well.

The naïve method preserves a feature of the original data—the sharp drop at the anniversary when many people coming off one-year contracts quit—that was lost in the sparse calculation.

1. Hi Michael, Interesting post!

1) In your survival example: Are you using the hazard definition that is truly a probability (a life table one)? Then converting to odds and then likelihood as shown? Then multiplying all these together for a given individual?

2) Interested in this form of likelihood and the naive bayes model. I'm used to seeing NB as below -- are they the same as what you are describing?

P(y=1|X) =

(P(X|y=1)*P(y=1))/ (P(x)) where P(x) is

product(P(x_i | y=1))*P(y=1) + product(P(x_i | y=0))*P(y=0).

Shame that the term naive has a negative connotation. You could brilliantly be arriving at a new approach and yet be ashamed of the patent! :P

2. Yes, curious as to where that definition of likelihood came from? Can you provide a reference? I have never seen it before.

3. Hi Steven,

Yes, I am using the life table method. Time is discreet and measured in days. The hazard probability estimate for each tenure is simply the ratio of the number of subscribers who have ever quit at that tenure divided by the number who ever achieved that tenure (and so were at risk for quitting).

There are lots of ways of writing down the naive Bayes model which all turn out to be the same. I wrote it in the form I thought was easiest to implement in excel.

I have a column for the overall hazard and a column for each conditional hazard. (One for each market, one for each rate plan, etc.) Each of the conditional hazard columns is also represented as a likelihood. This could be calculated as the ratio of the hazard given the condition to the hazard given not the condition, but another formulation is easier: the odds of quitting given the condition divided by the overall odds.

4. Hi Anonymous,

I actually gave the reference in the original post. It is Data Analysis Using SQL and Excel by Gordon Linoff. In Chapter 10, there is a good treatment of the relationship of probability, odds, and likelihood. It gives the more familiar definition of
likelihood(A|B)=P(B|A)/P(B|~A) but also points out the alternative definition I used.

The chapter has a nice little Venn diagram that I can't reproduce in a comment, but I can still use the numbers to illustrate that the two formulations yield the same result.

There are 100 customer, 40 of whom stop. So the overall probability of a stop is 40% and the odds of stopping are 40/60 or 2/3. Of the 40 who stop, 2 are in a particular market. There are 10 people total in the market, so 20% stop and the odds of stopping in the market are 2/8.

The likelihood that someone in the market will stop by the first method is the probability of being in the market given that they stopped (2/40) divided by the probability of being in the market given that they did not stop (8/60). That is 120/360 = 3/8.

The likelihood that someone in the market will stop by the second method is the odds of stopping in the market (2/8) divided by the overall odds of stopping (40/60). That is 120/360 = 3/8.

5. You mention logistic regression earlier in the article, but did you ever consider employing it to model the hazards, as a drop-in replacement for Naive Bayes? Especially given the sparseness of the data under certain parameters (like with Smallville-Mail-Top) it seems that the benefits of _not_ making the Naive Bayes assumption could be substantial.

On a related note, it's worth mentioning for purposes of completeness that leaving a term out of the product due to missing data is equivalent to multiplying by 1.0, which means the probability of the missing data field is actually being treated as 0.5. In other words, leaving the term out does encode an assumption about the distribution of the missing data. Fortunately, it's the very reasonable assumption that in the absence of other data, uniform probabilities should be assumed. But it's worth making explicit. It just so happens that doing the computations in “odds-space” somewhat obscures what’s going on. In “probability-space” a 0.5 would have to be inserted into the product to achieve the same effect, which would seem to make it more obvious that the uniform assumption is being made.

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