ABC for missing data
I received this email a few days ago:
I am an hard-core reader of your blog and thanks to your posts I have recently started being interested to ABC (and Bayesian methods in general). I am writing you to ask for suggestions on the application of the semi-automatic ABC à la Fearnhead & Prangle. The reason why I am contacting you instead of addressing the authors is because (i) you have been involved in the RSS reading of their paper and (ii) you are an authority on ABC, and therefore you are probably best suited and less biased on such issue. I am applying ABC with the semi-automatic statistics selection provided in Fearnhead and Prangle (2012) to a problem which can be formalized as a hidden Markov model. However I am unsure of whether I am making a huge mistake on the following point: let’s suppose we have an unobserved (latent) system state X (depending on an unknown parameter θ) and a corresponding “corrupted” version which is observed with some measurement error, e.g.
Y = X + ε,
where ε is the measurement error independent of X, ε is N(0, σ²), say. Now their setup does not consider measurement error, so I wonder if the following is correct. Since I can simulate n times some X’ from p(X|θ) am I allowed to use the corresponding “simulated” n corrupted measurements
Y’ = X’ + ε’
(where ε’ is a draw from p(ε|σ)) into their regression approach to determine a (vector of) summary statistic S=(S1,S2) for (θ,σ)? I mean the Y’ are draws from a p(y|X’,θ,σ) which is conditional on X’. Is this allowed? Wilkinson (2008) is the only reference I have found considering ABC with measurement-error (the ones by Dean et al (2011) and Jasra et al (2011) being too technical in my opinion to allow a practical implementation) and since he does not consider a summary statistics-based approach (e.g. Algorithm D, page 10) of course he is not in need to simulate the corrupted measurements but only the latent ones. Therefore I am a bit unsure on whether it is statistically ok to simulate Y’ conditionally on X’ or if there is some theoretical issue that makes this nonsense.
to which I replied
…about your model and question, there is no theoretical difficulty in simulating x’ then y’given x’ and a value of the parameters. The reason is that
.the proper marginal as defined by the model. Using the intermediate x’ is a way to bypass the integral but this is 100% correct!…
a reply followed by a further request for precision
Although your equation is clearly true, I am not sure I fully grasp the issue, so I am asking for confirmation. Yes, as you noticed I need a
y’ ~ f(y’|θ,σ)
Now it’s certainly true that I can generate a draw x’ from f(x’|θ,σ) and then plug such x’ into f(y’|x’,θ,σ) to generate y’. By proceeding this way I obtain a draw (y’,x’) from f(y’,x’|θ,σ). I think I understood your reasoning, on how by using the procedure above I am actually skipping the computation of the integral in:
Is it basically the case that the mechanism above is just a classic simulation from a bivariate distribution, where since I am interested in the marginal f(y’|θ,σ) I simulate from the joint density f(y’,x’|θ,σ) and then discard the x’ output?
which is indeed a correct interpretation. When simulating from a joint, the marginals are by-products of the simulation.