# SMC sampler weights when $K_n$ leaves $\pi_{n-1}$ invariant

I cannot seem to find this proof anywhere. Suppose I choose $$K_n$$ to leave $$\pi_{n-1}$$ invariant, and $$L_{n-1}$$ to be the reversal kernel. I want to show that the incremental weights $$w_n(x_{n-1}, x_n) = \frac{\pi_n(dx_n) L_{n-1}(x_n, dx_{n-1})}{\pi_{n-1}(dx_{n-1})K_n(x_{n-1}, dx_n)} = \frac{\pi_n(x_{n-1})}{\pi_{n-1}(x_{n-1})}.$$

# Attempt

### Invariance

$$K_n$$ leaving $$\pi_{n-1}$$ invariant means $$\pi_{n-1}K_n = \pi_{n-1}$$ This is known as "global balance".

### Reversibility

In MCMC context, usually one cannot verify global balance directly. Instead, one makes sure that $$K_n$$ is $$\pi_{n-1}$$-reversible, which is a much stronger condition. $$\pi_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n) = \pi_{n-1}(dx_n) K_n(x_n, dx_{n-1}).$$ This is known as "detailed balance", or "local balance". It is easy to show that detailed balance implies global balance.

### Reversal

$$L_{n-1}$$ being the reversal kernel of $$K_n$$ means $$\pi_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n) = \pi_n(dx_n) L_{n-1}(x_n, dx_{n-1})$$

### "Wrong" Proofs

1. Using only $$L_{n-1}$$ reversal: If I use directly the fact that $$L_{n-1}$$ is the reversal kernel, I get constant incremental weights $$w_n(x_{n-1}, x_n) = \frac{\pi_{n-1}(dx_{n-1})K_n(x_{n-1}, dx_n)}{\pi_{n-1}(dx_{n-1})K_n(x_{n-1}, dx_n)} = 1$$
2. Using $$L_{n-1}$$ reversal and reversibility: Perhaps, I'm thinking, people write $$K_n$$ leaving $$\pi_{n-1}$$ invariant, but they actually mean $$K_n$$ being $$\pi_{n-1}$$-reversible. In that case, using these two conditions together leads to $$w_n(x_{n-1}, x_n) = \frac{\pi_{n-1}(dx_n)K_n(x_n, dx_{n-1})}{\pi_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n)}$$ which seems wrong.
3. Feynman-Kac: Integrating $$x_{n-1}$$ out of the two-step Feyman-Kac formula we get an expression for $$\pi_n$$ $$\pi_n(dx_n) = \int_{x_{n-1}\in X} w_n(x_{n-1}, x_n) \pi_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n)$$ an alternative expression is given by the change of variables $$\pi_n(dx_n) = \frac{\pi_n(dx_n)}{\pi_{n-1}(dx_n)} \pi_{n-1}(dx_n)$$ Assuming $$K_n$$ leaves $$\pi_{n-1}$$ invariant we get \begin{align} \pi_n(dx_n) &= \frac{\pi_n(dx_n)}{\pi_{n-1}(dx_n)} \pi_{n-1}(dx_n) \\ &= \frac{\pi_n(dx_n)}{\pi_{n-1}(dx_n)} \pi_{n-1} K_n(dx_n) \\ &= \frac{\pi_n(dx_n)}{\pi_{n-1}(dx_n)} \int_{x_{n-1}\in X} \pi_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n) \\ &= \int_{x_{n-1}\in X} \frac{\pi_n(dx_n)}{\pi_{n-1}(dx_n)} \pi_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n). \end{align} where we can identify $$w_n(x_{n-1}, x_n) = \frac{\pi_n(x_n)}{\pi_{n-1}(x_n)},$$ which is different from what has been suggested in the literature.

### Only proof I could find

The only proof I could generate assumes $$K_n$$ leaves $$\pi_n$$ invariant, and not $$\pi_{n-1}$$. In that case, $$L_{n-1}$$ being the reversal of $$K_n$$ means $$\pi_n(dx_{n-1}) K_n(x_{n-1}, dx_n) = \pi_n(dx_n) L_{n-1}(x_n, dx_{n-1}).$$ Plugging this into the expression for the incremental weights lead to the correct incremental weight $$w_n(x_{n-1}, x_n) = \frac{\pi_n(x_{n-1})}{\pi_{n-1}(x_{n-1})}.$$