By "not a direct function of my parameters" I mean the following. I have some observed K-dimensional data and a model that can generate synthetic data based on 6 free parameters.

I use this model to generate synthetic data sets (by setting the 6 free parameters) and compare them with my observed data using the (log) Poisson likelihood rate as my likelihood:

$$Lkl = 2 \sum_i m_i - n_i + n_l ln\frac{n_i}{m_i}$$

where $i$ is the ith K-dimensional bin, $n_i$ is the number of observed data points in that bin, and $m_i$ is the number of synthetic data points in that bin. Each model generated by setting the 6-parameters to some value, will have its own $m_i$ distribution. This is what I mean with "my likelihood is not a direct function of my parameters".

I've been using Bayesian MCMC with some success but the convergence is really slow. I learned about Hamiltonian Monte Carlo, but this method requires the gradient of the log posterior, which means obtaining the gradient of my likelihood.

Can this method be applied to my problem at all?


1 Answer 1


It looks like you might be re-inventing Approximate Bayesian Computation (ABC). The core of ABC is to simulate many synthetic datasets and compare the synthetic data to the observed data based on some summary statistics. You can read Marin et al. (2012) for a review.

  • $\begingroup$ I am aware of ABC, but I'm not sure it is the answer in my case since: a) I know the shape of my likelihood, and b) it is really not that hard or costly to estimate it. My question was aimed at knowing if I can actually apply HMC to an intractable likelihood. $\endgroup$
    – Gabriel
    Commented Sep 6, 2018 at 18:05
  • $\begingroup$ My understanding is that you are actually using a pseudo-likelihood, which looks a lot like a summary statistic for ABC (or ABC-MCMC). In particular, $Lkl$ is not a deterministic function of your parameters, so you can't define a gradient for HMC. $\endgroup$ Commented Sep 7, 2018 at 7:39
  • $\begingroup$ I wasn't aware of the pseudo-likelihood concept, but what you say makes a lot of sense. I'll give ABC a try and see if it has better performance than the standard MCMC I'm using right now. Thank you Robin! $\endgroup$
    – Gabriel
    Commented Sep 7, 2018 at 15:16

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