In a Bayesian MCMC model, if we plug the average of posterior draws back into the Likelihood, would it be estimating the Posterior Predictive?

Suppose we have a Bayesian model with data $$y$$ and a parameter to be estimated, $$\theta$$. Then the posterior is written as:

$$p(\theta | y) \propto p(y|\theta)p(\theta)$$

Suppose that we used an MCMC to obtain draws of the posterior, $$p(\theta | y)$$. Suppose we then take the average of these draws and then plug the average back into the likelihood, $$p(y|\theta)$$, and plot it across the support on $$y$$. Would this distribution be approximating the posterior predictive? Or is it approximating something completely different?

The posterior predictive is approximated by the average of the sampling densities over the MCMC sample $$p(y|y^\text{obs})\approx\frac{1}{T}\sum_{t=1}^T p(y|\theta^{(t)})$$ To simulate from this distribution, one need pick one of the $$\theta^{(t)}$$ at random and simulate $$Y$$ from the corresponding $$p(y|\theta^{(t)})$$ [and not for all the $$\theta^{(t)}]$$.
• If I wanted to build the histogram of the simulated distribution, it seems there are two approaches. The first (yours), is randomly picking $\theta^{(t)}$ at random, simulate $Y$, iterate this many times, then plot the $Y$'s I get. The second, seems to be picking $\theta^{(1)}$, simulating $Y^{(1)}$, picking $\theta^{(2)}$, simulating $Y^{(2)}$, etc, until I get $T$ such $Y$'s and getting their histogram. This second one doesn't pick the $\theta^{(t)}$'s randomly, is it the same? What is the purpose of choosing $\theta^{(t)}$ randomly (which I also assume is by replacement)? – user321627 Feb 20 at 2:14
• If you want to draw/plot the approximate distribution, you simply need averaging the $p(y|\theta^{(t)})$'s, there is no need to draw/simulate $Y_i$'s. If you want to exactly draw/simulate from the approximate predictive [rhs], sampling with replacement is needed. – Xi'an Feb 20 at 9:44