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

Hi,

We can explain it cleanly using the standard HMM joint probability formula.

P(S, O) = P(s1) × ∏{t=2 to T} P(st | st−1) × ∏{t=1 to T} P(ot | st)

Now plug in the values from the problem:

P(s1) = 1/3

P(st | st−1) = 1/3 for all states

P(ot | st) = 1/3 for all states and observations

So:

P(S, O) = (1/3) × (1/3)^(T−1) × (1/3)^T

Combine exponents:

P(S, O) = (1/3)^(2T)

The key point is that this expression does not depend on the specific hidden state sequence S at all.

So every possible state sequence of length T gives the same probability.

Conclusion: there is no unique most likely hidden state sequence (no unique MLE). All state paths are equally likely explanations for the observation sequence.

Answer 2

Hello professor,

The most likely explanation is that there is no unique state sequence. Because all transition, emission, and initial probabilities are exactly 1/3, every possible path through the states is equally probable.

Mathematically, the joint probability for any sequence of length T=13 is:

P(S, O) = (Initial) * (Transitions) * (Emissions)

P(S, O) = (1/3) * (1/3)^12 * (1/3)^13 = (1/3)^26

Since this value is a constant, it is independent of the states chosen. This means the specific observations provide no information to favor one path over another; a sequence of 13 "Red" states is just as likely as any other combination. Every possible state sequence is a tie for the Maximum Likelihood Estimate.

Answer 3

Hi professor,

I want to point out that all probabilities here are uniform. Every state has the same initial probability of 1/3, every transition is 1/3, and every emission is 1/3. This means for any hidden state sequence of length 13, the joint probability works out the same way:

P = (1/3) x (1/3)^12 x (1/3)^13 = (1/3)^26

Since this value is identical for every possible sequence, there is no unique most likely explanation. All 3^13 possible hidden state sequences are equally likely, whether it is something like:

R, R, R, R, R, R, R, R, R, R, R, R, R

or:

R, G, B, R, G, B, R, G, B, R, G, B, R

or anything else.

Answer 4

All emission probabilities are equal:

P(observation | state) = 1/3 for every state and observation.

All transition probabilities are equal:

P(next state | current state) = 1/3 for all transitions.

Initial states are also equally likely:

P(R) = P(G) = P(B) = 1/3

Therefore,  given that any state sequence produces the observations with probability (1/3)^13, any sequence has transition probability (1/3)^12 and Initial probability is (1/3), total probability for any sequence is :

(1/3) × (1/3)^12 × (1/3)^13 = (1/3)^26

Thus, all possible hidden state sequences have exactly the same probability.There is no unique most likely explanation; all state sequences are equally likely.

Answer 5

Hello,

The most likely explanation is that any sequence of states is equally likely.

Reasons:

1.  Uniform Emissions: Every state (Red, Green, Blue) emits every observation (1, 2, 3) with the exact same probability 1/3. Therefore, the observations provide no information to distinguish which state you are in.

2.  Uniform Transitions: From any state, you transition to any other state (including staying) with the exact same probability 1/3.

3.  Uniform Initial State: You start in any state with equal probability 1/3.

Because every possible path through the states has the exact same mathematical probability, there is no single "most likely" sequence. All 3^13 possible state sequences are equally probable.