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

We use the standard HMM joint probability:

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

Given:
P(s1 = R) = 1
P(ot | st) = 1/3 for all states and observations

So the emission part is always:

∏ P(ot | st) = (1/3)^T

The difference now comes from the transition matrix:

R → R, G, B each with 1/3
G → G with 1/2, B with 1/2
B → B with 1

So unlike the previous case, transitions are NOT uniform anymore. This means the probability of a state sequence depends heavily on how many times we use G or B transitions.

Now we reason qualitatively:

• Staying in B is the most “stable” (probability 1 each time)
• Staying in G is next best (1/2 each time)
• R is the least stable (keeps splitting into 3 options with 1/3)

So the most likely hidden path will try to:

• move into B as quickly as possible
• then stay in B for as long as possible

Given the observation sequence is completely uninformative (all emissions are equally likely), the hidden state path is driven only by transitions.

Final conclusion:

The most likely explanation is a state sequence that starts at R (given), quickly transitions into B, and then stays in B for most or all of the remaining steps, because B → B has probability 1 which maximizes the transition likelihood.

Answer 2

Hello,

Since all states have identical emission probabilities (1/3) the observations do not help distinguish between them. The most likely sequence depends only on maximizing transition probabilities.

Starting in Red we should move to Blue immediately (p=1/3) because Blue is an absorbing state (p=1). Staying in Red or Green reduces the total probability while staying in Blue preserves it.

Most likely sequence:

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

Total Probability:

(1/3) * (1/3)^25 = (1/3)^26

Answer 3

Hi professor,

I noticed that the transition structure here is the key. Once we leave R we can never return to it, and once we reach B we stay in B forever. So the hidden state sequence always looks like some R's, then some G's, then B's for the rest.

Since all emissions are uniform (1/3 for each observation regardless of state), the emission probabilities do not help us differenciate between states. So the most likely sequence is the one that maximizes the transition probabilities.

Since B --> B with probability 1 (the highest possible transition probability), we want to reach B as soon as possible and stay there. From R, the fastest path to B is R --> B directly (p=1/3), or R --> G --> B, etc. Since staying in B is certain, transitioning to B earlier gives higher overall probability.

Therefore the most likely explanation is transitioning to B at the very first step and staying there for all 25 time steps:

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

Answer 4

All emission probabilities are equal:

P(1 | state) = P(2 | state) = P(3 | state) = 1/3

So the observations do not help us distinguish between R, G, and B. The answer depends only on transition probabilities.

Initial state is R. From R:

R-> R = 1/3

R-> G = 1/3

R -> B = 1/3

After reaching B:

B -> B = 1

So the best strategy is to move from R to B immediately, then stay in B forever.

Transition probability:

P(R -> B) × P(B-> B)^23

= 1/3 × 1^23

= 1/3

This is larger than going through R or G, because those transitions have probabilities less than 1. Therefore the most likely 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

Answer 5

Hello,

The most likely explanation (state sequence) is that the system starts in Red, transitions to Green at some point, and then eventually transitions to Blue, where it stays for the remainder of the sequence.

However, looking closely at the specific probabilities:

1.  Emissions are uniform 1/3 for all symbols in all states. This means the observations themselves provide no information to distinguish between states. The likelihood depends entirely on the transition probabilities.

2.  Transitions:

    *   From R: You can go to R, G, or B with equal probability 1/3.

    *   From G: You must go to G or B 1/2 each. You cannot go back to R.

    *   From B: You must stay in B p=1. This is an absorbing state.

Since emissions don't help, we want the path with the highest transition probability.

*   Staying in R forever has probability (1/3)^(T-1).

*   Moving to G and staying there has probability involving 1/3 for the first step and 1/2 for subsequent steps. Since 1/2 > 1/3, staying in G is more likely than staying in R.

*   Moving to B and staying there has probability involving 1/3 or 1/2 for the entry step, and 1 for all subsequent steps. Since 1 > 1/2 > 1/3, once you enter B, the probability cost for remaining steps is zero (log-probability is 0).

Therefore, the single most likely specific path is the one that enters the absorbing state Blue as early as possible, because staying in Blue has a probability of 1 per step, which is higher than staying in Red 1/3 or Green 1/2.

The earliest you can enter Blue is at step 2 R -> B.

Path: R, B, B, B, ..., B

Let's compare:

*   Path R, R, R...: Prob = (1/3)^24

*   Path R, G, G... Prob = 1/3 * (1/2)^23

*   Path R, B, B...: Prob = 1/3 * 1^23 = 1/3

Clearly, 1/3 is much larger than the others.

The most likely state sequence is:

Red, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue, Blue