I guess that the prior and posterior predictive distributions can be considered expectation of $p(y|\theta )$ (in case of prior predictive distribution) and $p(\widetilde{y}|\theta )$ (in case of posterior predictive distribution).
But, I cannot fully convince myself about the idea. This is because, I am not sure if $p(y|\theta )$ and $p(\widetilde{y}|\theta )$ are multiplied by correct probabilities.
Assuming that $p(\theta )d\theta $ and $p(\theta|y )d\theta $ are probability of getting $\Theta$, we multiply $ p(y|\theta )$ and $p(\widetilde{y}|\theta )$ by the probabilities respectively. If the two expressions of the predictive distributions actually calculated expectations, it would mean that we would be able to say, for example, "we get $ p(y|\theta )$ with probability of $p(\theta )d\theta $". Can we say this?(same question for the posterior predictive distribution) Whether I can say that or not is what I struggle to understand.
Since $ p(y|\theta )$ depends on $\theta $, so we could say it ?
The below is equations for the two distributions.
p() is a densintifunction.
Prior predictive distribution
$p(y) = \int p(y|\theta )p(\theta )d\theta $
Posterior predictive distribution ($\widetilde{y}$ is unobserved quantity while $y$ is observed)
$p(\widetilde{y}|y) = \int p(\widetilde{y}|\theta )p(\theta|y )d\theta $
