# SMC Samplers - Optimal Backward Kernel Explanation

In Sequential Monte Carlo Samplers of Del Moral (2006) we see that the optimal backward kernel is $$L_{n-1}^{\text{opt}} (x_{n-1} \mid x_n) = \frac{\eta_{n-1}(x_{n-1}) K_n(x_n \mid x_{n-1})}{\eta_n(x_n)}$$ I am very confused by the notation used in the paper. It seems like sometimes $$L$$ and $$K$$ are kernels, sometimes they are densities and same for $$\eta$$ and $$\pi$$.

• What does the numerator mean? Is it $$K_n$$ operating on the left on the measure $$\eta_{n-1}$$ $$\eta_{n-1} K_n(A) = \int_E \eta_{n-1}(d x_{n-1}) K_n(x_{n-1}, A) \qquad \qquad A\in\mathcal{E}$$ or is it just a multiplication of their densities? $$\frac{d \eta_{n-1}}{d \lambda} \cdot \frac{d K_n(x_{n-1}, \cdot)}{d\lambda}$$
• What does the denominator mean? I have never seen a ratio of measures like that. Doesn't seem to make any sense? Surely $$L_{n-1}: E\times\mathcal{E}\to [0,1]$$ is a Markov Kernel whereas both $$\eta_{n-1} K_n$$ and $$\eta_n$$ are measures? $$L_{n-1}^{\text{opt}}(x_{n}, ) = \frac{\eta_{n-1} K_n}{\eta_n}$$

This notation is used thoughout the whole paper and it's very confusing for me!

$$L_{n-1}^{\text{opt}} (x_{n-1} \mid x_n) = \frac{\eta_{n-1}(x_{n-1}) K_n(x_n \mid x_{n-1})}{\eta_n(x_n)}$$
should be interpreted as follows: the measure $$\eta_{n-1}(x_{n-1}) K_n(x_n \mid x_{n-1})$$ can be given an unambiguous meaning as a probability measure on the space $$E_{n-1} \times E_n$$ (one could emphasise this by writing it as e.g. $$\eta_{n-1}(dx_{n-1}) K_n(dx_n \mid x_{n-1})$$). The optimal $$L$$-kernel is then obtained by taking regular conditional probabilities of this joint measure with respect to $$x_n$$.
• Can you explain explicitly how it is obtained? I know that the unnormalized incremental weight is $$\tilde{\omega}_n(x_{n-1}, x_n) = \frac{\eta_n(dx_n) L_{n-1}(x_n, dx_{n-1})}{\eta_{n-1}(dx_{n-1})M_n(x_{n-1}, dx_n)}$$ and apparently the optimal backward kernel is the one that minimizes the variance of the weights. In my mind, minimial variance is zero variance and so we aim for a constant incremental weight. $$\tilde{\omega}_n(x_{n-1}, x_n) = \text{const}.$$ The incremental weight is a Radon-Nikodym derivative $$\frac{d (\eta_n \otimes L_{n-1})}{d(\eta_{n-1} \otimes K_n)}$$ Jun 30 at 17:37
• and by definition of RN derivative we know this means that the two measures are proportional up to a constant $c > 0$ $$\eta_n(dx_n) L_{n-1}(x_n, dx_{n-1}) = c \eta_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n).$$ Jun 30 at 17:39
• Intuitively, I understand that either side represents the joint distribution over $(x_{n-1}, x_n)$ and so by """conditioning""" I can guess that it is enough for me to first find the marginal $$\eta_n(dx_n) = \int_{x_{n-1}\in E} \eta_n(dx_n) L_{n-1}(x_n, dx_{n-1}) = c \int_{x_{n-1}\in E} \eta_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n)$$ and then divide the joint by this to obtain the desired result $$L_{n-1}^{\text{opt}}(x_n, dx_{n-1}) = \frac{\eta_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n)}\int_{x_{n-1}\in E} \eta_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n)}$$ Jun 30 at 17:44
• which can equivalently be written as $$L_{n-1}^{\text{opt}}(x_n, dx_{n-1}) = \frac{\eta_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n)}{\eta_n(dx_n)}$$ but I cannot seem to justify this with measure theory Jun 30 at 17:46