I am studying Hidden Markov Models and I'm trying to understand the following exercise:

Consider Hidden Markov Model with hidden states $h_{1:T} = \{h_1,...,h_T\}$ and observed states $v_{1:T}=\{v_1,...,v_T\}$.

Let $v_t\in\{0,1,2\}, h_t\in\{0,1\}$ and the parameters of the model are as follows. $p(h_1=1)=0.5$, the transition matrix $T=\begin{pmatrix} 0.5 & 0.8 \\ 0.5 & 0.2 \end{pmatrix}$ and the emission matrix $\begin{pmatrix}0.3 & 0.5 \\ 0.6 & 0 \\ 0.1 & 0.5 \end{pmatrix}$.

Question 1: Suppose that you observe the sequence of outcomes $v_{1:10}=[0,1,0,2,0,2,1,0,2,0]$. Is $h_1$ independent of $h_{10}$ in $p(h_{1:10}|v_{1:10})$ given this particular sequence?

Question 2: Compute the filtering distributions $p(h_8|v_{1:8})$ and $p(h_9|v_{1:9})$ (Hint: you do not need to run the full Forward algorithm here.).

Answer to question 1: We notice in the emission matrix that $p(v_t=1|h_t=1)=0$, so it is certain that in this particular sequence, $h_2=0,h_7=0$. Since the distribution $p(h_{1:10}|v_{1:10})$ is a Markov chain, $h_{1:t-1} \perp\kern-5pt\perp h_{t+1:n} | h_t$, so we have that in this particular sequence, we $p(h_{1:10}|v_{1:10}) = p(h_{1:10}|v_{1:10},h_2,h_7)$, making $h_1$ and $h_{10}$ independent.

Answer to question 2: By the previous question we know that $p(h_8|v_{1:10},h_7,h_2) = p(h_8|h_7=0)$. Hence, by the transition matrix, $p(h_8|h_7=0) = \begin{pmatrix} 0.5 \\ 0.5 \end{pmatrix}$ and for $p(h_9|v_{1:10})$:

\begin{align} p(h_9|v_{1:10}) & \propto p(v_9|h_9)\sum_{h_8}p(h_9|h_8)p(h_8|v_{1:8}) \\ & = \begin{pmatrix} p(v_9=2|h_9=0)\left(\frac{1}{2}p(h_9=0|h_8=0)+ \frac{1}{2} p(h_9=0|h_8=1)\right) \\ p(v_9=2|h_9=1)\left)\frac{1}{2} p(h_9=1|h_8=0)+\frac{1}{2}p(h_9=1|h_8=1)\right) \end{pmatrix} \\ & = \begin{pmatrix} 0.1(\frac{1}{2}\frac{1}{2}+0.8\frac{1}{2}) \\ \frac{1}{2}(\frac{1}{2}\frac{1}{2}+0.2\frac{1}{2}) \end{pmatrix} = \begin{pmatrix} 0.065 \\ 0.175 \end{pmatrix} \end{align}

After normalizing it becomes $0.27, 0.73$

Is my reasoning correct for these questions?



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.