Hidden Markov Model Solved Exercise

Question:

1. Mr. X is happy someday and angry on other days. We can only observe when he smiles, frowns, laughs, or yells but not his actual emotional state. Let us start on day 1 in the happy state. There can be only one state transition per day. It can be either to happy state or angry state. The HMM is shown below;

Assume that q_t is the state on day t and o_t is the observation on day t. Answer the following questions;

(a) What is P(q₂ = Happy)?

(b) What is P(o₂ = frown)?

(c) What is P(q₂ = Happy | o₂ = frown)?

(d) What is P(o₁ = frown o₂ = frown o₃ = frown o₄ = frown o₅ = frown, q₁ = Happy q₂ = Angry q₃ = Angry q₄ = Angry q₅ = Angry) if π = [0.7, 0.3]?

Solution:

(a) What is P(q₂ = Happy)?

The question is to find the probability of Mr. X is Happy on day 2. It is given that the first day he was Happy. For the second day, we need to find the transition probability P(q₂ = Happy | q₁ = Happy).

P(q₂ = Happy | q₁ = Happy) = 0.8

(b) What is P(o₂ = frown)?

We need to find the probability of observation frown on day 2. But we don’t know the states whether he is happy or not on day 2 (we know he was happy on day 1). Hence, the probability of the observation is the sum of products of observation probabilities and all possible hidden state transitions.

P(o₂ = frown) = P(o₂ = frown | q₂ = Happy) + P(o₂ = frown | q₂ = Angry)

            = P(Happy | Happy)* P(frown | Happy) + P(Angry | Happy)* P(frown | Angry)

            = (0.8 * 0.1) + (0.2 * 0.5) = 0.08 + 0.1 = 0.18

(c) What is P(q₂ = Happy | o₂ = frown)?

Here, we need to find the probability of hidden state on day 2 as Happy given the observation on that day as frown. This conditional probability cannot be calculated directly. Hence, we apply Bayes’ rule to solve as follows;

P(q₂ = Happy | o₂ = frown) = (P(o₂ = f| q₂ = H) * P(q₂ = H)) / P(o₂ = f)

                                                = (P(Happy | Happy)* P(frown | Happy)) / 0.18

[Note: 0.18 is taken from the answer for question (b)]

                                                = (0.8 * 0.1) / 0.18 = 0.08 / 0.18 = 0.4444

(d) What is P(o₁ = frown o₂ = frown o₃ = frown o₄ = frown o₅ = frown, q₁ = Happy q₂ = Angry q₃ = Angry q₄ = Angry q₅ = Angry) if π = [0.7, 0.3]?

Here, we need to find the probability of the observation sequence “frown frown frown frown frown” given the state sequence “Happy Angry Angry Angry Angry”. π is the initial probabilities.

P(f f f f f, H A A A A)  

            = P(f|H)*P(f|A)*P(f|A)*P(f|A)*P(f|A)*P(H)*P(A|H)*P(A|H)*P(A|H)*P(A|H)

            = 0.1 * 0.5 * 0.5 * 0.5 * 0.5 * 0.7 * 0.2 * 0.2 * 0.2 * 0.2

            = 0.000007

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Related Links:

• Go to Hidden Markov Model Formal Definition page

• Go to Natural Language Processing (NLP) home

• Go to NLP Glossary - Dictionary page

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