(I guess stats.SE is the right place for this)

I'm reading Albert's book "Bayesian computation with R". To get theprior predictive density, he extensively uses this formula

$$f(y) = \frac{f(y\mid \lambda) g(\lambda)}{g(\lambda | y)}, \qquad (1)$$

where $f(y\mid \lambda)$ is the (whatever model we have; in the specific example 3.3, Poisson) sampling density on $y$, $g(\lambda)$ prior density on parameter $\lambda$, and $g(\lambda \mid y)$ posterior density on $\lambda$.

Now, I'm thoroughly confused about this! Okay, the equation $$ f(y\mid \lambda) g(\lambda)) = g(\lambda \mid y) f(y) $$ is just the familiar Bayes' theorem rearranged, but I can't quite explain why we're using it, because don't we in the first place find the posterior by Bayes' theorem $$g(\lambda | y) = \frac{g(y\mid \lambda) g(\lambda)}{g(y)}$$ (or so, I'm not exactly sure about the notational logic behind $g$s and $f$s)? Why don't we end up with $f(y) = f(y)$?

I thought predictive distributions (prior and posterior) were solely attained by 'marginalizing', you know, this thing:

$$f(y) = \int_\Theta f(y\mid \theta) g(\theta) d \theta \qquad (2) $$

edit. (for prior predictive; and for posterior predictive something like $f(\tilde{y} \mid y) = \int_\Theta f(\tilde{y}\mid \theta) g(\theta \mid y) d \theta $)


The formulas$$f(y) = \frac{f(y\mid \lambda) g(\lambda)}{g(\lambda | y)}$$ for the prior predictive (or marginal) density and $$g(\lambda | y) = \frac{f(y\mid \lambda) g(\lambda)}{f(y)}$$for the posterior density are both derived from the joint distribution$$f(y\mid \lambda) g(\lambda)) = g(\lambda \mid y) f(y)$$which uses $f$ for densities of samples and $g$ for densities of parameters. They are simply used in different settings. For instance the prior predictive is useful to assess the fit of a given model.

  • $\begingroup$ Thank you! While your answer clarifies a bit, it doesn't answer all. for example, what's the difference between (1) and (2) (see edited question)? $\endgroup$
    – kekkonen
    May 23 '15 at 13:19
  • $\begingroup$ no difference at all. $\endgroup$
    – Xi'an
    May 23 '15 at 13:20

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