From the article of wikipedia


I see that one generate samples from the proposal $\pi(x_k^{(L)}\vert x_{o:k-1}^{(L)},y_{1:k})$, however, the role of $y_{1:k}$ is not clear to me.

Let's assume that we have the following functional forms: $y_t=ax_t+\epsilon_t$ and $x_t=bx_{t-1}+\eta_t$

Then the system equation density given by $p(x_k^{(L)}\vert x_{k-1})$ can be implemented in R for a normal distribution as $rnorm(1,x_k^{(L)}-bx_{k-1}^{(L)},\sigma_2)$. Now, if we assume as well a normal distribution for the observation equation I think that one could simulate from the proposal as $rnorm(1,y_k-ax_k^{(L)},\sigma_1)$ but I'm not sure whether my interpretation is correct or not. In addition, it seems that we would have to generate samples from the proposal first in order to compute the density $p(x_k^{(L)}\vert x_{k-1})$.

I would appreciate any hint you may give.


1 Answer 1


For this problem, you must take into account the entire information available, which means the full conditional distribution of $x_k$ (the $(L)$ index is not necessary for the distributions) given $x_{0:(k-1)}$ and given $y_{1:k}$, which simplifies into the full conditional distribution of $x_k$ given $x_{k-1}$ and given $y_{k}$ in the noisy AR model you consider here. Hence, \begin{align*}p(x_k|x_{0:(k-1)},y_{1:k})&=p(x_k|x_{k-1},y_{k})\\ &\propto p(x_k|x_{k-1})\times p(y_k|x_k)\\ &=\varphi(\{x_k-bx_{k-1}\}\sigma_1^{-1})\times\varphi(\{y_k-ax_{k}\}\sigma_2^{-1})\\ &\propto \varphi(\{x_k-[\sigma_1^{-2}bx_{k-1}+\sigma_2^{-2}ay_{k}]/[\sigma_1^{-2}+\sigma_2^{-2}]^{-1}\}\times[\sigma_1^{-2}+\sigma_2^{-2}]^{1/2})\end{align*}

  • 1
    $\begingroup$ would the likelihood in the third line be $\varphi(\{y_k-ax_k\}\sigma_2^{-1})$ instead? $\endgroup$
    – nopeva
    Commented Jan 25, 2015 at 10:32

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