Body: <p>I am going through <a href="https://ieeexplore.ieee.org/document/18626" rel="nofollow noreferrer">Rabiner 1989</a> and he writes that the discrete probability density function of duration <span class="math-container">$d$</span> in state <span class="math-container">$i$</span> (that is, staying in a state for duration <span class="math-container">$d$</span>, conditioned on starting in that state) is <span class="math-container">$$p_i(d) = {a_{ii}}^{d-1}(1-a_{ii})$$</span></p>
  
  <p>(<span class="math-container">$a_{ii}$</span> is the state transition probability from state <span class="math-container">$i$</span> to state <span class="math-container">$i$</span> - that is, staying in the same state).</p>
  
  <p>He then continues to say that the expected durations in a state, conditioned on starting in that state, is <span class="math-container">$$\overline d_i = \sum_{i=1}^\infty d p_i(d) = \frac{1}{1-a_{ii}}$$</span></p>
  
  <p>Where does the coefficient <span class="math-container">$d$</span> (in <span class="math-container">$\sum_{i=1}^\infty d p_i(d)$</span>) come from?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-02T07:44:40.723Z
Comments.Text: Looks like a typo and sum should be over $d$, not $i$, in the second equation. But I am not familiar with that tutorial.
Comments.UserDisplayName: Neil Slater
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-02T20:08:54.163Z
Comments.Text: @NeilSlater Still, even if it is summing over $d$, what does the coefficient $d$ represent? Is it adding $p_i(d)$ each time it stays in state $i$?
Comments.UserDisplayName: jstycrpsc
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-02T22:04:23.243Z
Comments.Text: You already state that: duration 
Comments.UserDisplayName: Neil Slater
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-03T04:54:54.273Z
Comments.Text: @NeilSlater Sorry I was not clear. Why do we multiply $p_i(d)$ by $d$?
Comments.UserDisplayName: jstycrpsc
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-03T06:47:29.463Z
Comments.Text: . . . .because it is taking a mean/expected value by multiplying each possible value by its probability of occurring
Comments.UserDisplayName: Neil Slater
Tags: <math><probability><hidden-markov-model>
Title: Expected duration in a state
UserDisplayName: jstycrpsc
ThreadText: # Expected duration in a state
  
  (Asked by: username_0 on 2020-03-02)
  
  I am going through [Rabiner 1989](https://ieeexplore.ieee.org/document/18626) and he writes that the discrete probability density function of duration $d$ in state $i$ (that is, staying in a state for duration $d$, conditioned on starting in that state) is $$p\_i(d) = {a\_{ii}}^{d-1}(1-a\_{ii})$$
  
  ($a\_{ii}$ is the state transition probability from state $i$ to state $i$ - that is, staying in the same state).
  
  He then continues to say that the expected durations in a state, conditioned on starting in that state, is $$\overline d\_i = \sum\_{i=1}^\infty d p\_i(d) = \frac{1}{1-a\_{ii}}$$
  
  Where does the coefficient $d$ (in $\sum\_{i=1}^\infty d p\_i(d)$) come from?
  
  **Tags:** math, probability, hidden-markov-model
  > **Comments:**
  > **username_1**: Looks like a typo and sum should be over $d$, not $i$, in the second equation. But I am not familiar with that tutorial.  
  > **username_0**: @NeilSlater Still, even if it is summing over $d$, what does the coefficient $d$ represent? Is it adding $p_i(d)$ each time it stays in state $i$?  
  > **username_1**: You already state that: duration   
  > **username_0**: @NeilSlater Sorry I was not clear. Why do we multiply $p_i(d)$ by $d$?  
  > **username_1**: . . . .because it is taking a mean/expected value by multiplying each possible value by its probability of occurring  
  
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  ## 0 answers
  
  