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!


1 Answer 1


In the setting where everything has densities, I think that the interpretation is fairly standard, so I will not focus on that case here.

At the measure-theoretic level, the numerator

$$L_{n-1}^{\text{opt}} (x_{n-1} \mid x_n) = \frac{\eta_{n-1}(x_{n-1}) K_n(x_n \mid x_{n-1})}{\eta_n(x_n)}$$

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 $x_n$.

  • $\begingroup$ 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 $$ \frac{d (\eta_n \otimes L_{n-1})}{d(\eta_{n-1} \otimes K_n)} $$ $\endgroup$ Commented Jun 30, 2023 at 17:37
  • $\begingroup$ and by definition of RN derivative we know this means that the two measures are proportional up to a constant $c > 0$ $$ \eta_n(dx_n) L_{n-1}(x_n, dx_{n-1}) = c \eta_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n). $$ $\endgroup$ Commented Jun 30, 2023 at 17:39
  • $\begingroup$ 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 $$ L_{n-1}^{\text{opt}}(x_n, dx_{n-1}) = \frac{\eta_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n)}{\displaystyle \int_{x_{n-1}\in E} \eta_{n-1}(dx_{n-1}) K_n(x_{n-1}, dx_n)} $$ $\endgroup$ Commented Jun 30, 2023 at 17:44
  • $\begingroup$ 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 $\endgroup$ Commented Jun 30, 2023 at 17:46

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