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I am trying to conceptualize the analytical relationship between the prior distribution and posterior distribution obtained by MCMC for Bayesian inference. Sorry for the non-rigorous notation but I hope it still communicates the idea.

This thesis by Zhou suggests that the posterior $\Pr(X^n) = \Pr(X^n | \textit{data})$ of parameter $X$ can be obtained through $n$ acceptance steps via the prior $\Pr(X^0)$ and product of transition probabilities $\Pr(X^t | X^{t-1}) = \Pr(X^t | X^{t-1}, \textit{data})$: \begin{equation} \Pr(X^n) = \Pr(X^0) \prod_{t=1}^n \Pr(X^t | X^{t-1}) \end{equation} ($\textit{data}$ is omitted to simplify notation.)

Asher et al. writes that the limiting distribution that approximates the posterior distribution is obtained by: \begin{equation} \Pr(X^n) = \lim_{n\to\infty} \Pr(X^n | X^0) \end{equation}

According to the Chapman-Kolmogorov equation, \begin{align} \Pr(X^n | X^0) = \int_{-\infty}^{\infty} \Pr(X^n | x) \Pr(x | X^0) dx \end{align}

My question is: I am not sure this relationship equivalent? \begin{align} \Pr(X^n | X^0) \stackrel{?}{=} \Pr(X^0) \prod_{t=1}^n \Pr(X^t | X^{t-1}) \end{align}

Is there another way to conceptualize the relationship between $\Pr(X^0)$ and $\Pr(X^n)$ through transition probabilities?

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  • $\begingroup$ This is most confusing as you only mention a sequence $(X_0,X_1,\ldots,X_n)$ and no parameter or observation. In which sense is this Bayesian or related with MCMC? $\endgroup$
    – Xi'an
    Jun 24, 2018 at 5:40
  • $\begingroup$ Thanks for the comment - I've now stated that $X$ is in fact the parameter and the observation or data is implied but not included to simplify the notation. $\endgroup$
    – hatmatrix
    Jun 24, 2018 at 19:09
  • $\begingroup$ Still confused by the fact that the parameter is of increasing size. What is the meaning of $n$? $\endgroup$
    – Xi'an
    Jun 25, 2018 at 5:20
  • $\begingroup$ $n$ is the index of iteration - while it looks like an exponent I have seen this notation to describe steps in MCMC. $\endgroup$
    – hatmatrix
    Jun 25, 2018 at 9:03

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The equation \begin{equation} \Pr(X^n) = \Pr(X^0) \prod_{t=1}^n \Pr(X^t | X^{t-1}) \end{equation} is incorrect, it should be \begin{equation} \Pr(X^n) = \int \Pr(x^0) \prod_{t=1}^{n-1} \Pr(x^t | x^{t-1})\Pr(X^n | x^{n-1}) \,\text{d}x^0\cdots\,\text{d}x^{n-1} \end{equation}or \begin{equation} \Pr(X^n|X^0) = \int \prod_{t=1}^{n-1} \Pr(x^t | x^{t-1})\Pr(X^n | x^{n-1}) \,\text{d}x^1\cdots\,\text{d}x^{n-1} \end{equation}

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    $\begingroup$ This clears it up! Thanks - so basically you have to integrate between all states x$^1$ to x$^{n-1}$. So is it correct that the product should go from $t=1$ to $t=n-1$ rather than $t=1$ to $t=n$? $\endgroup$
    – hatmatrix
    Jun 26, 2018 at 12:37
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    $\begingroup$ Sorry, typo corrected! $\endgroup$
    – Xi'an
    Jun 26, 2018 at 15:10

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