Here's the problem, as far as I understand it. For given measure spaces:
$$(\mathcal{X}_0, B(\mathcal{X}_0), \mathbb{P}_0)$$
$$(\mathcal{X}_1, B(\mathcal{X}_1), \text{don't-care-about-this-measure-yet?})$$
Chopin and Papaspiliopoulos, in their 2021 Sequential Monte Carlo book, introduce the idea of a probability kernel:
$$P_1(x_0, dx_1)$$
where $x_0 \in \mathcal{X}_0$, and $dx_1 \in B(\mathcal{X}_1)$. This kernel can do a lot of nice things for us. For example, it defines a measure over the product space $(\mathcal{X}_0 \times \mathcal{X}_1, B(\mathcal{X}_0) \times B(\mathcal{X}_1))$:
$$\mathbb{P}_1(dx_{0:1}) = \mathbb{P}_0(dx_0)P_1(x_0, dx_1)$$
Which should be interpreted as the integral
$$\mathbb{P}_1(A) = \int_{A}\mathbb{P}_0(dx_0)P_1(x_0, dx_1)$$
Ok, I think I understand what's going on here. What's tripping me up is what comes next. The book tells us:
Section 4.1 decomposed the joint distribution $P(dx_{0:1})$ into the marginal at time $0$ and the conditional given in terms of the kernel. However, we can decompose the distribution in a “backwards” manner instead: $$\mathbb{P}_1\left(\mathrm{~d} x_0\right) P_1\left(x_0, \mathrm{~d} x_1\right)=\mathbb{P}_1\left(\mathrm{~d} x_1\right) \overleftarrow{P}_0\left(x_1, \mathrm{~d} x_0\right)$$ where the equality is understood as one of two joint distributions
(4.1) should be in reference to the decomposition I provided above, which I'll now restate:
$$\mathbb{P}_1(dx_{0:1}) = \mathbb{P}_0(dx_0)P_1(x_0, dx_1)$$
Ok. Now I'm totally lost. I think a good question to help narrow this down for me is this:
- Why is the LHS decomposition here written in terms of $\mathbb{P}_1(dx_0)$ instead of $\mathbb{P}_0(dx_0)$?