MLE for given HMM and Observation Sequence

Question

We are given an MDP with 3 states: Red, Green, Blue (RGB). 

Emission probabilities:  

R --> 1 (p=1/3), 2 (p=1/3), 3 (p=1/3)

G --> 1 (p=1/3), 2 (p=1/3), 3 (p=1/3)

B --> 1 (p=1/3), 2 (p=1/3), 3 (p=1/3)

Transition Matrix

R --> R (p=1/3), G (p=1/3), B (p=1/3)

G --> R (p=1/3), G (p=1/3), B (p=1/3)

B --> R (p=1/3), G (p=1/3), B (p=1/3)

These are the observations: 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 1, 2, 2

What is the most likely explanation for this observation sequence?
Initial state is not known and is equally likely from all states.

Answer 1

Without solving this problem mathematically, I think, it is very obvious that the most likely explanation for this observation sequence is about their equality as emission probabilities and values of transition matrices are the same. Therefore, there is no unique MLE; any state sequence of the same length is equally probable. 

If we solve this problem using the mathematical equations:

P(S,O)=P(s1​) ∏( P(st​ ∣ st−1​) ∏ ​P(ot​ ∣ st​)

P(s1​)=1/3

​P(st∣st−1)= 1/3

​ P(ot∣st)=1/3​

After putting the values into the equation, we get :

P(S, O)=(1/3​) ^ 2T, which means P(S, O) is independent of S, so whichever state sequence is there, they all have the same likelihood. 

Answer 2

From given question, we can see that all emission probabilities are equal and all transition probabilities are equal, so every possible hidden state sequence has exactly the same probability. We can write it as below:

• P(R)=P(G)=P(B)=1/3     (Initial state probabilities)
• P(1∣R)=P(2∣R)=P(3∣R)=1/3         (Emission probabilities) 
• P(R∣R)=P(G∣R)=P(B∣R)=1/3       (Transition probabilities)​​

So for any hidden-state sequence of length 13:

P(States, observations) = P(s₁) ∏¹³_t=1 P(o_t ∣ s_t) ∏¹³_t=2 ​P(s_t ​∣ s_t-1), or simply

P(states,observations) = P(s₁​) * P(o₁​∣s₁​) * P(s₂​∣s₁​) * P(o₂​∣s₂​) * P(s₃​∣s₂) * P(o₃​∣s₃​).....P(s₁₃∣s₁₂) * P(o₁₃​∣s₁₃​)

P(states,observations)= (1/3)²⁶ 

So, all hidden-state sequences are equally likely. For example:

1) B,G,R,B,R,G,G,B,R,R,G,B,B

2) G,G,G,G,G,G,G,G,G,G,G,G,G

3) R,G,B,R,G,B,R,G,B,R,G,B,R

all have the same probability. 
Therefore, there is no single most likely state sequence, as every possible RGB state sequence of length 13 is equally likely.

Answer 3

Since all probabilities are same all states are symmetric. This means that any RGB sequence of length 13 is equally likely. Probability for each is: (1/3)^26