# HMM path probabilities: three complete numerical examples

Subject: Natural Language Processing

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/nlp/#tee/formulas/f-hmm/hot-cold-paths

Canonical: https://deep-learning-study-room.vercel.app/read/nlp/tee/formulas/f-hmm--hot-cold-paths/

Formulas explained through full worked examples

The following model uses hidden weather states Hot (H) and Cold (C). Observations are numbers 1, 2 and 3. These numbers are observation labels, not quantities to multiply directly. Their emission probabilities come from the table.

| Initial state | Probability |
| --- | --- |
| H | 0.8 |
| C | 0.2 |

| From → to | H | C |
| --- | --- | --- |
| H | 0.7 | 0.3 |
| C | 0.4 | 0.6 |

| Emission state | Observation 1 | Observation 2 | Observation 3 |
| --- | --- | --- | --- |
| H | 0.2 | 0.4 | 0.4 |
| C | 0.5 | 0.4 | 0.1 |

**Question. An HMM has π(H)=0.8, π(C)=0.2; H→H=0.7, H→C=0.3, C→H=0.4, C→C=0.6. Emissions for observations (1,2,3) are H:(0.2,0.4,0.4) and C:(0.5,0.4,0.1). Find the joint probability of hidden path H,C,H and observations 1,3,1.**

#### Given

Hidden path H→C→H. Observations 1→3→1. Model probabilities are specified in the question. There is no END factor.

#### To find

P(H,C,H; 1,3,1).

#### Formula and why it applies



$$
P=\pi_H b_H(1)a_{HC}b_C(3)a_{CH}b_H(1)
$$



There are three hidden states and observations: three emissions, two transitions, and one initial probability.

#### Solution

##### Step 1. Start Hot and emit 1



$$
0.8\times0.2=0.16
$$



##### Step 2. Transition Hot to Cold



$$
0.16\times0.3=0.048
$$



##### Step 3. Cold emits 3



$$
0.048\times0.1=0.0048
$$



##### Step 4. Transition Cold to Hot



$$
0.0048\times0.4=0.00192
$$



##### Step 5. Hot emits the last 1



$$
0.00192\times0.2=0.000384
$$



**Final answer.** Joint probability = 0.000384 = 3.84×10⁻⁴.

**Question. For π(H)=0.8, π(C)=0.2; H→H=0.7, H→C=0.3, C→H=0.4, C→C=0.6; emissions H:(0.2,0.4,0.4), C:(0.5,0.4,0.1) for labels (1,2,3), find the joint probability of path H,H,H and observations 1,3,1.**

#### Given

State path H→H→H, observations 1→3→1; no END factor.

#### To find

P(H,H,H;1,3,1).

#### Formula and why it applies



$$
P=\pi_H b_H(1)a_{HH}b_H(3)a_{HH}b_H(1)
$$



The state remains Hot; each transition still contributes its probability 0.7.

#### Solution

##### Step 1. Initial state and first emission



$$
0.8(0.2)=0.16
$$



##### Step 2. First self-transition



$$
0.16(0.7)=0.112
$$



##### Step 3. Emit observation 3



$$
0.112(0.4)=0.0448
$$



##### Step 4. Second self-transition



$$
0.0448(0.7)=0.03136
$$



##### Step 5. Emit final observation 1



$$
0.03136(0.2)=0.006272
$$



**Final answer.** Joint probability = 0.006272. A self-transition is not automatically probability 1.

**Question. For π(H)=0.8, π(C)=0.2; transitions H:(0.7,0.3), C:(0.4,0.6) in next-state order H,C; emissions H:(0.2,0.4,0.4), C:(0.5,0.4,0.1) for observations 1,2,3, find the score of path C,C,H with observations 1,3,1.**

#### Given

Hidden path C→C→H. Observation sequence 1→3→1. No end-state factor.

#### To find

The joint path-and-observation probability.

#### Formula and why it applies



$$
P=\pi_C b_C(1)a_{CC}b_C(3)a_{CH}b_H(1)
$$



The initial state is Cold, so use π(C), not π(H). Each emission is selected from the state at the same position.

#### Solution

##### Step 1. Start Cold and emit 1



$$
0.2(0.5)=0.1
$$



##### Step 2. Stay Cold



$$
0.1(0.6)=0.06
$$



##### Step 3. Emit 3 from Cold



$$
0.06(0.1)=0.006
$$



##### Step 4. Move Cold to Hot



$$
0.006(0.4)=0.0024
$$



##### Step 5. Emit 1 from Hot



$$
0.0024(0.2)=0.00048
$$



**Final answer.** Joint probability = 0.00048.

**Yaad rakho:** Har column par: state ka emission. Columns ke beech: transition.