lag decided by the user, which depends on the forgetting properties of the model. By marginalisation of the empirical smoothing distribution $\widehat{p}_{\theta}(x_{1:\kappa_t}|y_{1:\kappa_t})$ over $x_{1:t-2}$ and $x_{t+1:\kappa_t}$, we obtain the approximation $$\begin{aligned}
    \widehat{p}_{\theta}^\Delta(\dn x_{t-1:t}| \mathbf{y}) 
    \triangleq 
    \sum_{i=1}^N 
    w_{\kappa_t}\pIdx{i} 
    \delta_{\tilde{x}_{\kappa_t,t-1:t}\pIdx{i}}
    (\dn x_{t-1:t}).
    \label{eq:FL2step}\end{aligned}$$ Here, we use the notation $\tilde{x}_{\kappa_t,t}^{(i)}$ to denote the ancestor at time $t$ of particle $x_{\kappa_t}^{(i)}$ and $\tilde{x}_{\kappa_t,t-1:t}^{(i)} = \{ \tilde{x}_{\kappa_t,t-1}^{(i)}, \tilde{x}_{\kappa_t,t}^{(i)} \}$. Inserting – into provides an estimator of , $$\begin{aligned}
    \widehat{\mathsf{S}}_T(\theta|u)
    &=
    \nabla \log p(\theta) +
    \sum_{t=1}^{T}
    \sum_{i=1}^N    
    w_{\kappa_t}\pIdx{i}
    \xi_{\theta} \Big( \tilde{x}_{\kappa_t,t}\pIdx{i}, \tilde{x}_{\kappa_t,t-1}\pIdx{i} \Big),
    \label{eq:FisherScoreParticleApproximation}\end{aligned}$$ which is used in the proposal distributions in .

Estimation of the Hessian
-------------------------

The negative Hessian of the parameter log-posterior can be written as $$\begin{aligned}
    \mathsf{I}_T(\theta) = -\DDtheta \log p(\theta) - \DDtheta \log p_{\theta}( \mathbf{y} ),
        \label{eq:defSecondOrder}\end{aligned}$$ where it is assumed that the Hessian of the log-prior $\DDtheta \log p(\theta)$ can be calculated analytically. The negative Hessian of the log-likelihood, also known as the *observed information matrix*, can using *Louis’ identity* [@CappeMoulinesRyden2005] be expressed as $$\begin{aligned}
    - \DDtheta \log p_{\theta}(\mathbf{y}) 
    &= 
    \Dtheta \log p_{\theta}(\mathbf{y})^2 
    -
    \mathbb{E}_{\theta} 
    \Big[ \DDtheta \log p_{\theta}(\mathbf{x},\mathbf{y}) \Big| \mathbf{y} \Big]    \nonumber \\
    &-\mathbb{E}_{\theta} \Big[ 
    \Dtheta \log p_{\theta}(\mathbf{x},\mathbf{y})^2
    \Big| \mathbf{y} \Big].
    \label{eq:LouisIdentity}\end{aligned}$$ Here, we have introduced the notation $v^2=vv^{\top}$ 