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I am running a fully Bayesian MCMC procedure to estimate some time series models, and my model has a lot of parameter estimates. In particular, one of these parameters, $\phi$, is $\in [-1,1]$. The conditional density is nonstandard, so I have implemented an independance-chain Metropolis-Hastings with a truncated normal $N(\hat{\phi},\sigma^{2}_{\phi})1(|\phi|<1)$ where the mean and variance are based on some other parameters. I was a little unsure how reliable my model was, so I have run some simulated data through it where I have fixed $\phi=0.50$.

I have a simulated sample size of $500$ a loop size of $20,000$, and a burn-in period of $5,000$ and a computation time of around 60 minutes (and yes my matrices are sparse).

Surprisingly (or maybe not), the posterior is skewed to the left and the posterior mean/median are well below $0.50$ (typically around $0.4$). Obviously, when the sampler gets close to one, it gets pushed to the left and can wander off to the left near negative values. I do not believe this is "bad", but the mean and median won't do a good job here in summarising $\hat{\phi}$. The mode looks like a much more sensible estimate, but the creditability intervals are wide (too wide).

Wikipedia thinks using the posterior mode is a bad idea: http://en.wikipedia.org/wiki/Maximum_a_posteriori_estimation

Is there a usual treatment for summarising the posterior when dealing with a constrained parameter space?

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    $\begingroup$ I don't understand the question. Are you asking what is the best way to summarize a posterior distribution using a single value? Why do you want to do that in the first place? $\endgroup$
    – Tom Minka
    Commented Nov 28, 2014 at 15:51
  • $\begingroup$ Please check the edit. I need to summarise it as I am conducting a simulation study to validate the authenticity of my model $\endgroup$
    – akkp
    Commented Nov 28, 2014 at 21:36
  • $\begingroup$ How do you know that the mean won't do a good job? $\endgroup$
    – Tom Minka
    Commented Nov 28, 2014 at 22:48
  • $\begingroup$ I've done that (as I wrote above). The posterior distributions are skewed to the left - apologies if this was not clear from the question $\endgroup$
    – akkp
    Commented Nov 29, 2014 at 13:16
  • $\begingroup$ The posterior median has the property that it is equally likely to be above or below the true $\phi$, when $\phi$ is sampled from the prior. It sounds like you want this property to be true for a given fixed value of $\phi$. However, that is a frequentist property, not a Bayesian one. $\endgroup$
    – Tom Minka
    Commented Nov 30, 2014 at 22:28

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To increase accuracy near the boundaries of the parameter space, a common approach is to make a variable transformation $\theta = f(\phi)$, where $f$ maps $[-1,1]$ into $[-\infty,\infty]$, compute the posterior mean of $\theta$, then transform this estimate back to $\phi$. This is equivalent to using Bayesian decision theory where the loss function emphasizes errors near the boundary of the original parameter space.

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  • $\begingroup$ Thanks. This makes sense. I suppose that a suitable transform would be $\frac{e^{x}}{1-e^{x}}$. How would I go about actually doing this? How does the Jacobian come into play here? Suppose my posterior was normal. How does this change the implementation? Thanks Tom! $\endgroup$
    – akkp
    Commented Dec 3, 2014 at 11:37

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