How can a future observation being predicted even if under sample independency (posterior predictive distribution)?

I was reading https://people.stat.sc.edu/Hitchcock/stat535slidesday18.pdf and I came across with:

It gives us an way to get a distribution for a new observation called for $$x_{new}$$ given the past sample values $$\mathbf{x}$$ but $$x_{new}$$ independs from $$\mathbf{x}$$ so $$P(x_{new}|\mathbf{x})=P(x_{new})$$ and:

So here are my questions:

1-Why does $$P(x_{new}|\mathbf{x})=\int_{\Theta}P(x_{new}|\theta,\mathbf{x})P(\theta|\mathbf{x})d\theta$$ hold?

2-If so then Why is use it worth since I'm able to evaluate $$P(x_{new}|\mathbf{x})$$ by the second figure?

$$p(x_{new}|x)$$ is the posterior predictive distribution. It can be computed as an integral over the parameter $$\theta$$ as stated in your question, in fact this is just an application of marginal probability density functions (https://en.wikipedia.org/wiki/Marginal_distribution): $$p(x_{new}|x)=\int_\theta p(x_{new},\theta|x) d\theta$$.

$$p(x_{new})$$ is the prior predictive distribution. $$p(x_{new}|x)$$ is never equal to $$p(x_{new})$$ except in trivial cases, because $$x_{new}$$ is not independent of $$x$$. When the slides mention independence of $$x$$ and $$x_{new}$$, they actually mean independence conditional on $$\theta$$ i.e. the only 'connection' between $$x$$ and $$x_{new}$$ is via the parameter $$\theta$$.

Conditional independence (https://en.wikipedia.org/wiki/Conditional_independence): in this context it means that $$p(x_{new}|x,\theta)=p(x_{new}|\theta)$$. This is a standard assumption that is made in Bayesian modelling. It's particularly clear that this is reasonable if, say, the data are (conditionally) independent Bernoulli trials with parameter $$p$$. If we know $$p$$, then knowing the outcome of any number of Bernoulli trials ($$x$$) won't tell us anything more about the outcome of future trials ($$x_{new}$$) (because they're conditionally independent). If we don't know $$p$$, then $$x$$ can tell us a lot about $$x_{new}$$ (because they're not independent).

If conditional independence doesn't hold, we can still proceed but the computations might be more complex. This might happen if, for example, $$x$$ comprises observation of a continuous time stochastic process over a time interval $$[0,T]$$ and $$x_{new}$$ is observations over the future interval $$(T,T+1]$$.

• What do you mean by 'independence conditional on θ' and why** is that true or even reasonable? Commented Nov 18, 2021 at 23:39
• I made some edits... Commented Nov 18, 2021 at 23:42
• it sounds like $x$ is an useless information if we know $\theta$ and it pretty makes sense.There is just a problem remaining, I didn't stated $P(x_{new}|x)=\int_{\theta} p(x_{new},\theta|x)d\theta$ I do $P(x_{new}|x)=\int_{\theta} p(x_{new}|\theta ,x)p(\theta|x)d\theta$ and your link just tell us about two random variables and its marginalization not 3 as we are in. (It actually tell us a little bit in the end's page but it doesnt use conditional regarding.) Commented Nov 19, 2021 at 0:01
• The two integrands in your comment are equal. This is because $P(A,B|C)=P(A|B,C)P(B|C)$ Commented Nov 19, 2021 at 0:12
• great, thank you Commented Nov 19, 2021 at 0:23