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
**************
Related Links:
- Go to Hidden Markov Model Formal Definition page
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