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 --> G (p=1/2), B (p=1/2)

B --> B (p=1)

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

What is the most likely explanation for this observation sequence?
Initial state is R.

Answer 1

Since all states have the same emission probabilities, the observations do not change which state path is best.

So we only compare the transition probabilities. As stated, initial state is fixed, so:
s1 = R

From R we can go to R (p=1/3), G (p=1/3), B (p=1/3), which means that all are equal for the first move. But after we reach B, then we will stay in B as B -> B (p = 1). 

Thus, the most likely state sequence is R followed by B for all remaining observations, because moving to B immediately avoids any further transition probability loss. 

So path is: R,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B,B

The full probability including the observations for this path is P = (1/3)²⁵ * 1/3 = (1/3)²⁶ 
The transitional probability for this path will be 1/3 * 1 * 1 *1 ... * 1 = 1/3

Answer 2

Since all emission probabilities are equal(1/3), the observations do not affect the choice of states. Therefore, we only need to maximize the transition probabilities. 

Starting from R, we compare options:

• Staying in R gives a probability 1/3 at each step
• Going to G gives at most 1/2 per step
• Going to B allows staying in B with probability 1 (since B = 1)

Thus, the optimal strategy is to move to B and remain there, because multiplying by 1 preserves the highest probability.

MLE for this observation:

R B B B B B B B B B B B B B B B B B B B B B B B B B 

Answer 3

We have a system where emissions are completely uninformative: every state (R, G, B) produces 1, 2, or 3 with equal probability $1/3$. This means the observation sequence does not help us distinguish between states at all. So the entire problem is driven only by the transition structure and the initial condition S1​=R.

We should focus on the transition rules, because that is where all the “meaning” is in this model:

• From R, you can go to R, G, or B with equal probability $1/3$. So R is a flexible starting state.
• From G, you either stay in G with probability $1/2$ or move to B with probability $1/2$. Importantly, once you leave G, you can never go back to R.
• From B, you stay in B forever with probability 1. So B is an absorbing “final lock” state.

So the structure is essentially directional: R → G → B, with increasing restriction as you move forward.

Since emissions give no information, the best explanation is the hidden path that has the highest transition probability structure.

That means the system prefers:

• Staying in a state rather than switching unnecessarily
• Using higher-probability self-transitions when available
• Avoiding early entry into the absorbing state B unless it is strongly supported.

The process starts in R (given), remains in R for some time, then eventually transitions into G. Once in G, it tends to stay there for a while due to the relatively strong self-loop probability. After that, it may transition into B, after which it becomes permanently stuck.

Even though the observed numbers look random, the hidden system behaves like a machine that gradually “loses flexibility”:

• R = free exploration stage
• G = semi-stable stage
• B = locked final stage

So the most likely explanation is not driven by the observations, but by the fact that the system naturally drifts forward and becomes more constrained over time.
A path that begins in R, spends a meaningful amount of time in R, then moves into G where it may linger, and eventually transitions into B, after which it remains fixed there.

Answer 4

Because all states have the same emission probabilities:

P(1∣R)=P(1∣G)=P(1∣B)=13

Because and same for observations 2 and 3, the observation sequence does not help distinguish between R, G, B. So, the MLE depends only on the transition probabilities. 

Initial state: q₁ = R , the possible next states are:

 R->R = 1/3,

R->G = 1/3,

R->B = 1/3.

But once we reach B:

B->B = 1

So the best strategy is to move to B as early as possible, because after that there is no transition penalty. Therefore: q₁ =R. Then choose: R->B

After that: B->B->B->B-> .......

For 25 observations, this gives 25 hidden states. Probability:

P = (1/3)²⁵ ⋅ (1/3) = (1/3)²⁶ 

Final answer: R, B, B, B, B, ...

Answer 5

Hi, 

Since all emission probabilities are equal, the observations do not help distinguish states.

So we only maximize the transition probability.

Initial state is R.

Best choice:

R -> B has probability 1/3
B -> B has probability 1 every time after that

So the most likely hidden-state sequence is:

R, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B

Why?

Its transition probability is:

(1/3) * 1 * 1 * … * 1 = 1/3

Any sequence that stays in R or goes through G has smaller probability because it keeps multiplying by 1/3 or 1/2.

Final answer:

R, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B