# Non-conventional form for a full conditional

I have a full-conditional of the form : $$p(X|...) \propto \exp(-(aX^2 +bX +c/X)),$$ All of the other full-conditionals of my model simplify to a simpler form and for a sake of simplicity, I would like to instantiate my model using Jags or WinBugs (which means that I need to explicit a form for $p(X|...)$ and that I cannot use explicitly a Metropolis step for $p(X|)$ (I think)).

Is it possible to do this? If so, how?

• Could you tell us anything about the likely values of $a$, $b$, $c$, and $X$? For some ranges of these values, this function is very close to Gaussian (but in others it is actually bimodal).
– whuber
Nov 7, 2012 at 16:36
• Is the whole term over $X$ or simply $c$? In the first case, it is an inverse Gaussian distribution. Nov 7, 2012 at 20:59
• whuber: I am investigating for typical values. Xi'an: I do not understand what you mean. can you please develop ? Nov 8, 2012 at 8:32

• I must be missing something about Jags (I guess it is the same for BUGS). Let's simply say my model is : $$p(y_i|X_i=x_i,A=a) \propto \exp(-(ax_i^2 +ax_i +a/x_i)),$$ with $$X_i \sim U(1,10)$$ $$A \sim U(0,10)$$ I do not know how to infer $p(A,(X_i)|(Y_i))$ in Jags naturally (I thought it is not possible). How can we do ? Thanks in advance. Nov 20, 2012 at 12:01
• Now I am confused: why do you want sampling a random variable from distribution, which does not depend on that random variable? Did you wanted to write as follows? $$p(X_{i}|a) \propto exp(-(ax_{i}^{2}+ax_{i}+a/x_{i}))$$ In this case, winbugs code for defining this distribution by zero trick can be found here: users.aims.ac.za/~mackay/BUGS/Manuals/Tricks.html where $L[i]=-(ax_{i}^{2}+ax_{i}+a/x_{i})$ Nov 20, 2012 at 12:29