1
$\begingroup$

HMM model

This is the HMM model considered in the question Emission probability matrix

And this is the emission probabilities for the respective states. There are two emission values, bringing an umbrella and not bringing an umbrella.

Description of the question:

The question considered the initial state as sunny. On day 2, the person brought an umbrella. On day 3, the person didn't carry an umbrella.

We have to find the probability that it's foggy on 3rd day.

Given: Prior probability of the caretaker carrying an umbrella is 0.5.

According to me, the answer to this question was addition of the terms in the red circles.

Where α1 = 0.8 * 0.1, α2 = 0.05 * 0.8, α3 = 0.15 * 0.3

Problem I faced: But the answer included dividing the answer I got by (0.5)2. That is, I guess they are dividing by the prior probability two times for the two transitions, but I'm not clear as to why they are dividing. Answer

$\endgroup$
0
$\begingroup$

Let $s_2,s_3$ be the states on the second and third days, and let $x_2,x_3$ be the indicators of carrying an umbrella on these days. Then $$ \newcommand{\Sunny}{\mathit{Sunny}} \newcommand{\Rainy}{\mathit{Rainy}} \newcommand{\Foggy}{\mathit{Foggy}} \newcommand{\Yes}{\mathit{Yes}} \newcommand{\No}{\mathit{No}} \Pr[s_3 = \Foggy \mid s_1 = \Sunny, x_2 = \Yes, x_3 = \No] = \frac{\Pr[s_3 = \Foggy, x_2 = \Yes, x_3 = \No \mid s_1 = \Sunny]}{\Pr[x_2 = \Yes, x_3 = \No \mid s_1 = \Sunny]} $$ It seems you might have forgotten the denominator.

$\endgroup$
  • $\begingroup$ Got your point! Thanks! $\endgroup$ – Amey Meher May 15 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.