Tutorial 9

1) Viterbi, Forward, and Backward Algorithms


Let $x$ be the sequence of symbols and let $\pi$ be the sequence of states sampled from a hidden Markov model.

Question 1: The Viterbi, forward, and backward algorithms can all be applied to hidden Markov models to produce tables. What does the final computed column of each table represent?

Viterbi algorithm:  

Forward algorithm:  

Backward algorithm:  







2) Forward Algorithm



$A = $
H A O
H -0.5 -1.6 -1.6
A -2.3 -1.2 -0.5
O -1.2 -1.6 -0.7
$E =$
S T
H -0.4 -1.2
A -1.6 -0.2
O -0.7 -0.7

The weather this week was $x = (T, T, T, S)$. We want to find $P(x) = \sum\limits_\pi P(x,\pi)$.

Question 2: Using the forward algorithm, complete the table below to find $P(x)$. Assume that we start in each state with probability $a_0 = 1/3$ ($A_0 = $ ln 1/3 = -1.1). Round to 1dp.

The top cell from column 1 was calculated as $F_H(1) = E_H(T) + A_{0,H} = -1.2 - 1.1 = -2.3$

The top cell from column 2 was calculated as $F_H(2) = E_H(T) + log\{ exp(-2.3 - 0.5) + exp(-1.3 - 2.3) + exp(-1.8 - 1.2) \} = -1.2 - 2.0 = -3.2$

The middle cell from column 2 was calculated as $F_A(2) = E_A(T) + log\{ exp(-2.3 - 1.6) + exp(-1.3 - 1.2) + exp(-1.8 - 1.6) \} = -0.2 - 2.0 = -2.2$

Start T T T S
0 0
H $-\infty$ -2.3 -3.2
A $-\infty$ -1.3 -2.2
O $-\infty$ -1.8 -2.0














What is $P(x)$?  






3) Backward Algorithm


$A = $
H A O
H -0.5 -1.6 -1.6
A -2.3 -1.2 -0.5
O -1.2 -1.6 -0.7
$E =$
S T
H -0.4 -1.2
A -1.6 -0.2
O -0.7 -0.7

Question 3: Using the backward algorithm, complete the table below to find $P(x)$. Assume that we have not modelled end states ($a_{k0} = 1$, $A_{k0} = $ ln 1 = 0). Round to 1dp.

The top cell from column 3 was calculated as $B_H(3) = log \{ exp(A_{HH} + E_H(S) + B_H(4)) + exp(A_{HA} + E_A(S) + B_A(4)) + exp(A_{HO} + E_O(S) + B_O(4)) \}$

$= log \{ exp(-0.5 + -0.4 + 0) + exp(-1.6 + -1.6 + 0) + exp(-1.6 + -0.7 + 0) \} = -0.6$


The middle cell from column 3 was calculated as $B_A(3) = log \{ exp(A_{AH} + E_H(S) + B_H(4)) + exp(A_{AA} + E_A(S) + B_A(4)) + exp(A_{AO} + E_O(S) + B_O(4)) \}$

$= log \{ exp(-2.3 + -0.4 + 0) + exp(-1.2 + -1.6 + 0) + exp(-0.5 + -0.7 + 0) \} = -0.8$


Start T T T S End
0 0
H $-\infty$ -0.6 0 $-\infty$
A $-\infty$ -0.8 0 $-\infty$
O $-\infty$ -0.7 0 $-\infty$














How do you compute the value in the top left cell ($B_{0}(0)$)?  

What is $P(x)$?  

Should this be the same as the number calculated in Question 2?  






4) Bayesian Inference


$$ P(\theta|D) = \frac{P(D|\theta) P(\theta)}{P(D)} $$


$D$ is the data observed and $\theta$ is our model/hypothesis.

Question 4a: If we observe symbol sequence $x = (T,T,T,S)$ and we believe that $\pi = (A,O,O,H)$, then what is $D$ and what is $\theta$? Substitute these terms into the Bayesian equation above.

D:  

$\theta$:  

$P(\theta | D) = $  



$$ P(\pi_i = k | x) = \frac{f_i(k) b_i(k)}{P(x)} $$


We can use the forward and backwards algorithms to calculate the probability of observing state k on the $i_\text{th}$ turn, $P(\pi_i = k | x)$, using the above equation.

Question 4b: Referring to the tables we made in Questions 2 and 3, what is the most probable state on the 1st day, given that the weather was $x = (T,T,T,S)$?

$P(\pi_1 = \text{happy} | x) = $  

$P(\pi_1 = \text{angry} | x) = $  

$P(\pi_1 = \text{okay} | x) = $  

Maximum posterior estimate: