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## Saturday, May 2, 2020

### 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 qt is the state on day t and ot 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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# Solved exercise in HMM, Exercise with solutions in hidden markov model, solutions to exercises in natural language processing

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