I have a question regarding section 7.1.1 here. I have two questions to ask you.
we shall be drawing values for $C_T,C_{T-1},\cdots,C_1$ in order.
1. The sentence
we shall be drawing values for $C_T,C_{T-1},\cdots,C_1$ in order.
means that the vector $C^{(T)}=(C_1,\ldots,C_T)$ is generated as
until$$C_{1}\sim p(C_{1}|C_{2},x^{(T)})$$
2. Simulating $C_T$ proceeds from the marginal relation $$\text{Pr}(C_TT | x^{(T)}, θ) ∝ α_T (C_T )$$ then simulating $C_{T-1}$ conditional on $C_T$ and $x^{(T)}$ is based on the backward conditional \begin{align}p(C_{T-1}|C_T,x^{(T)}, θ)&\propto \text{Pr}(C_{T-1} | x^{(T-1)}, θ) \text{Pr}(x_T,C_T| x^{(T-1)}, C_{T-1}, θ)\\ &\propto \text{Pr}(C_{T-1} | x^{(T-1)}, θ)\text{Pr}(C_T|C_{T-1}, θ)\text{Pr}(x_T|C_T,\theta)\\ &\propto\alpha(C_{T-1})\text{Pr}(C_T|C_{T-1}, θ) \end{align} by virtue of the hidden Markov representation. And so on for the next or rather previous terms in the series. Since each term only depends on higher order indices and known values $x^{(T)}$ and $\theta$, it is possible to simulate $C_T$, then $C_{T-1}$, then ... down to $C_1$. In the Gibbs sampler, this provides the conditional simulation of $C^{(T)}$.
C[20]<-sample(C("c1","c2","c3"),1,prob = A[T,]) Assume C[20]= c2. Then to compute C[19] we first compute the following 3 probabilities: Prob=A[19,'c1']*p(c1,c2),A[19,'c2']*p(c2,c2), A[19,'c3']*p(c3,c2) And then C[19]<-sample(C("c1","c2","c3"),1,prob = Prob)
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