0

The problem in hand is that the prior distribution which I have received from experts (loan recovery data) ranges from 0 to 100%. Thus a beta distribution was assumed. Where as the actual data shows that loan recovery can be more than 100% due to fees and interest charges. Thus the likelihood function does not follow a beta distribution simply because the values may be greater than 100%. Trying to use conjugate prior method to combine distributions to arrive at the posterior distribution. First I thought the beta-beta conjugate prior method will suffice. Rescaling the likelihood function to 0% - 100% does not seem correct. What do I do. Please help

1
  • Your data are non-negative ratios, so take the logarithm and model on R. A non-parametric method will probably serve you well. You might need to mix in a point mass at 0% (on the original scale, corresponding to on the log scale).
    – Cyan
    Commented Sep 26, 2013 at 18:08

1 Answer 1

2

The gamma distribution is conjugate and has support from 0 to . There is probably a way to convert beta parameters to gamma parameters analytically, but it might be quicker just to fit your beta prior to a gamma distribution. Here's an R example, assuming your beta prior is α=2,β=5:

require(MASS)
y <- rbeta(10000, 2, 5)
hist(y, breaks=31, freq=FALSE)
fit <- fitdistr(y, "gamma")
fit
x <- seq(0, 1, 0.01)
points(x, dgamma(x, fit$estimate[1], fit$estimate[2]))

You'll get output like this:

   shape         rate   
 2.61543039   9.18351169   
(0.03488704) (0.13501986)

You should be able to plug those values into your model with a gamma prior and accommodate values > 1.

1
  • Dear Eldan: This looks to be a simple solution, but could you please let me know of any supporting documents for the same? i have tried wikipedia and few other docs, but there is no indication of gamma-gamma conjugate prior model. May be I am wrong in understanding the text, please do guide me to some good reading material on the same. Thanks a ton for your help.
    – Bik
    Commented Oct 8, 2013 at 6:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.