I'm having a conceptual problem applying Markov Chain Monte Carlo for Bayesian parameter estimation; I know how MCMC works in principle but it's applying it to this problem that troubles me.

The theory has a function $K_1 \sin^2\theta + K_2 \cos^2\theta$ with parameters $K_1,K_2$ which predicts something (a particle decay width) based on an angle.

What I am given is the result of 10 experiments. Each experiment contains the following data:

  • Exact $K_1,K_2$ that were used to simulate experiment, as well as number of particle decays
  • Extracted $K_1$ and $K_2$ using moment analysis
  • Covariance matrix of $K_1$ and $K_2$

The goal is to estimate $K_1$ and $K_2$, the extracted ones being unnormalised, so they are independent and in a way there is a third parameter, the normalisation. Now, for MCMC I need a probability $P(D|x)$ of finding data $D$ given parameters $x$, and it's in determining this that I'm at a loss.

I see this is a trivial example where we could just compute a mean or something else and be done with it, but the goal is to use MCMC. And I'm not sure what $D$ in particular is in this case.


1 Answer 1


I have since worked out an approach. One option is to simply take,

$$\chi^2=[\vec D - \mathbb E(\vec D)]^T \Sigma^{-1}[\vec D - \mathbb{E}(\vec D)]$$

where $\vec D$ would be the measured values of $K_i$ and $\mathbb E(\vec D)$ would be the actual $K_i$ values. Then we can define a likelyhood based on $\exp(-\chi^2)$.

MCMC then proceeds as usual, picking a proposal to generate $K_i$ and a prior on $K_i$.

  • $\begingroup$ Sounds like you got the hang of it, but just to make things explicit: by choosing the likelihood to be $exp(-\chi^2)$, you have specified Gaussian error. $\endgroup$ Dec 30, 2022 at 19:41

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