Is it possible to further simplify the following expression (you can assume that $p$ are probability densities)
\begin{align} p(y_1, \dots, y_n \mid x_1, \dots, x_n) &= \frac{ p(y_1, \dots, y_n, x_1, \dots, x_n)}{p(x_1, \dots, x_n)}\\ &= \frac{\prod_{i=1}^n p(y_i, x_i)}{p(x_1, \dots, x_n)} \\ &= \frac{\prod_{i=1}^n p(y_i \mid x_i) p(x_i)}{p(x_1, \dots, x_n)}\\ &= \frac{\prod_{i=1}^n p(y_i \mid x_i) \prod_{i=1}^n p(x_i)}{p(x_1, \dots, x_n)} \end{align}
where the only assumption made is that we have a sequence of $n$ i.i.d. $(X_i, Y_i)$ random variables, so $p(x_1, \dots, x_n, y_1, \dots, y_n) = \prod_{i=1}^n p(x_i, y_i)$?
I have already tried different things, but I don't really know how to proceed. I think we can still change the denominator to $\int \dots \int p(x_1, \dots, x_n, \hat{y}_1, \dots, \hat{y}_n) d \hat{y}_1 \dots d \hat{y}_n$, right? But, even if that's right, then how do I proceed? The formula above should simplify to $\prod_{i=1}^n p(y_i \mid x_i)$, but I don't really know how that can be done, unless we assume that $X_i$ is independent of $X_j$, for $i \neq j$. But how come that follow from the fact that the tuples $(X_i, Y_i)$ are i.i.d.. I don't see how that follows. Some sources, like this one, say that joint independence implies marginal independence, but I am not sure if this applies here and how to prove actually. I would need to prove that, if $p(a, b, c, d) = p(a, b)p(c, d)$, then $p(a, c) = p(a)p(c)$ (and I guess $p(a, d) = p(a)p(d)$, $p(b, c) = p(b)p(c)$ and $p(b, d) = p(b)p(d)$).
Perhaps a different way to look at your denominator:
\begin{align*} p(x_1,\ldots,x_n) &= \int_\mathbb R \cdots \int_\mathbb R p(x_1,\ldots,x_n,y_1,\ldots,y_n)\,\mathrm dy_1\cdots \mathrm dy_n \\ &= \int_\mathbb R \cdots \int_\mathbb R \prod_{i=1}^n p(x_i,y_i)\,\mathrm dy_1\cdots \mathrm dy_n \\ &= \prod_{i=1}^n\int_\mathbb R p(x_i,y_i)\,\mathrm d y_i \\ &= \prod_{i=1}^n p(x_i). \end{align*}
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