I am a bit confused how to solve a Bayesian statistics problem. I have a parameter $\epsilon^s$ which is defined as following:

$$\epsilon^s=\frac{\epsilon-g(\pi,z)}{1-g^*(\pi,z)\epsilon}$$ where $g(\pi,z)$ has an analytical description (model) and it is a function of $z$ and $\pi$, but the empirical discrete probability of $z$ can be estimated, $p(z)$. The probability of $\epsilon^s$ is given by $$p_{\epsilon^s}(\epsilon^s)=\frac{1}{2\pi\sigma^2[1-\exp(-\frac{1}{2\sigma^2})]}\exp\left(-\frac{|\epsilon^s|^2}{2\sigma^2}\right)$$ and I assume $p_{\epsilon}(\epsilon|g)\equiv p_{\epsilon^s}(\epsilon^s|g)$. Now I would like to compute the likelihood of probability of measured $\epsilon$, i.e. $\epsilon^{m}=\epsilon+\epsilon^{err}$, for the given $g$ $$p_{\epsilon^m}(\epsilon^m|g)=\int p_{\epsilon}(\epsilon|g)p_{\epsilon^{err}}(\epsilon-\epsilon^m)d\epsilon$$

assuming a Gaussian probability distribution for the measurement error of $\epsilon$ ($\epsilon^{err}$). The likelihood for a set of $n$ measured $\epsilon$ and for a $g$ quantity with $\pi$ free parameters of the model is given by


My questions:

  1. What is the best way to compute $p_{\epsilon^m}(\epsilon^m_i|g_i)$ assuming both probabilities are Gaussian?
  2. how could I include the probability of $p(z)$ in the calculations and marginalize over?
  3. Can MCMC be a good method to solve this problem?

I will appreciate for any help!

  • $\begingroup$ Your model does not make sense as described. How do you come up directly with a pdf on $\epsilon^s$ rather than $\epsilon$? why do you have this extra-term in the normal pdf on $\epsilon^s$? why would the conditionals on $\epsilon$ and $\epsilon^s$ be the same given the fist equation? how is this conditional $p_{\epsilon}(\epsilon|g)$ defined? $\endgroup$
    – Xi'an
    Dec 28, 2014 at 21:05
  • $\begingroup$ @Xi'an the probability distribution of $\epsilon$ and $\epsilon^s$ are related with $p_{\epsilon}(\epsilon|g)\equiv p_{\epsilon^s}(\epsilon^s(\epsilon|g))|\frac{d^2\epsilon^s}{d\epsilon^2}|$. If I assume $g<<1$, then the Jacobian determinant is unity and it leads to the equality of $p_{\epsilon}(\epsilon|g)\equiv p_{\epsilon^s}(\epsilon^s|g)$. $\endgroup$
    – Dalek
    Dec 28, 2014 at 22:30
  • $\begingroup$ @Xi'an well, parameter $\epsilon^s$ describes an intrinsic property of an object, which its pdf is given by this quasi-Gaussian function. However $\epsilon$ is the observable quantity of this property of an object which has been distorted with $g$. $\endgroup$
    – Dalek
    Dec 28, 2014 at 22:40
  • $\begingroup$ The quasi Gaussian function is not a density, i.e. it does not integrate to one (1)... $\endgroup$
    – Xi'an
    Dec 29, 2014 at 9:28

1 Answer 1


If $\epsilon$ is a scalar (which it seems to be), then your integral is a 1D convolution of $p_\epsilon(\epsilon|g)$ with a Gaussian. If $p_\epsilon$ is fairly well behaved, you could evaluate it on a 1D grid and use any old signal processing toolbox to calculate the convolution (you could also quite easily code it by hand).

You could also include $p(z)$ in the same way. The neat thing about working in one or two dimensions is that brute force is often a feasible approach.

  • $\begingroup$ $\epsilon$ is a spin-2 tensor. Indeed it is a convolution but how could I compute the integral. $\endgroup$
    – Dalek
    Dec 28, 2014 at 22:37

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