If you are confused about the stationary distribution in Markov chain, my new Markov chain video might be helpful to understand it. This video tries to answer the question: Under what condition does Stationary Distribution exist?
Hope this is helpful for those needed.
I have a Markov chain problem below, where :
The problem
An urn initially contains 3 black balls and 1 red ball. The balls are indistinguishable to the touch. One ball is randomly drawn.
- If this ball is black, it is removed.
- If this ball is red, we put it back in the urn.
- The operation is repeated until the urn contains only the red ball.
1- Let Xn be the number of black balls contained in the urn after n prints.
Show that (Xn)nβN is a Markov chain.
2- Give its associated and reduced graph and its transition matrix. Is the chain homogeneous?
3- Classify the states.
4- What type of chain (absorbent, irreducible, ergodic) is it?
5- Give a possible trajectory of size 10.
6- Determine the expectation of the number of prints until the urn contains only the red ball.
7- Does the chain admit a stationary distribution? a borderline distribution? Calculate limP^(n), where 'n' tends towards infinity.
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Solution
1- We denote that E=(1,2,3) and the process is discrete to a discrete state space the state of Xn+1 depends only on the previous state Xn,
Therefore Xn is a Markov chain.
2-
- Transition Matrix :
- Associated graph :
The chain is not homogeneous because p(2/3) is different from p(2/1).
3- Classify the states: state classification: we have three transitory states (class):
- State 1 does not communicate with any state other than itself.
- State 2 communicates with state 1 but state 1 does not communicate (same for state 3 ).
Hence: E={1}U{2}U{3}.
4-
- The chain is not ergodic because it does not admit a recurrent state.
- The chain is not irreducible because it admits 3 classes.
- The chain is absorbent because state 1 is an absorbent state and all non-absorbent states (2/3) reach state 1.
5- Trajectory of size 10 :
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Help
- Can you please
Link: https://link.springer.com/article/10.1007%2FBF02018448
DOI: 10.1007/BF02018448
Published: January 1982
Couldn't find on Sci-hub or Libgen. I really need this for important work. Many thanks in advance!
Somewhat old paper, https://arxiv.org/abs/1506.04696. I recently spent some time going over this, and the paper has some great proofs and discussion. I did however,find myself looking at their theorem and thinking "how on earth did they find that form of the drift coefficient", and found the proof to be mainly about showing if you use that form, the result holds. That's not too enlightening if you want insight into how they found the result, so I went the other way myself, and in 1D it turns out to be somewhat straight-forward. I wrote it up if anybody else finds this view interesting
I am new to RL and what have a doubt regarding policy gradient theorem.
Why does there exists a stationary state distribution in policy gradient theorem ? i.e
why this turns to be a constant.(refer 13.2 in the below link)
I know it's the existence of the stationary state distribution that we do not take the derivative of the state distribution, and are able to take the derivative of the RL objective only using the derivation of the policy being learned.
To be more clear I am referring to the Policy Gradient theorem 13.2 in Sutton's latest version.(http://incompleteideas.net/book/bookdraft2017nov5.pdf)
I'm reading about Diffusion Maps Spectral Clustering. On page 3, the author is discussing the various interpretations of the first (largest) eigenvalue of the random-walk normalized Laplacian Similarity matrix (D^(-1)W) when using the Gaussian kernel for calculating similarity.
He specifically writes that the largest eigenvector "has a dual interpretation. The first is the stationary probability distribution on the graph, while the second is that Ο(x) [the x^(th) index of the eigenvector] is a density estimate at the point x. Note that for a general shift invariant kernel K(xβy) and for the Gaussian kernel in particular, Ο is simply the well known Parzen window density estimator."
I can't find anything else online about the relationship between kernel density estimation and the stationary distribution of a Markovian random walk on the data. Anybody seen this before, and/or can verify I'm understanding this equality correctly? It's a neat relationship between interpretations if it's true, and a somewhat new perspective on kernel density estimation for me.
I am reading about Regular Markov Chains from various sources, and am getting pretty confused by the various naming schemes used in different sources.
For example "Regular" refer to a property, both of the Markov Chain model itself, as well as its transition matrix. A Markov Chain is Regular if its state distribution is Steady-State, Stationary, or in Equilibrium, depending on what source you are reading from.
This vocabulary seems needlessly complex. How did this varied vocabulary grow, and why hasn't there been an effort to "consolidate" the statistics literature to a unified vocabulary?
Let X_n be a MC, P not regular
Say we have a stationary dist (pi_0, ..., pi_n) and P(X_0 = i) = 0.2, does this say anything?
To be more clear:
I ask because Karlin says when a stationary dist is not a limiting dist, P(X_n = i) is dependent on the initial distribution. What does this exactly mean?
Say I have a particular stationary distribution g(x) that I would like any initial distribution to converge to over time in the presence of diffusion. Is it possible to calculate a potential U(x) that makes this happen?
What is the reasoning and implication for making A(t) = A(t-1)?
A Markov chain on states 0, 1, ... has transition probabilities Pij = 1/(i+2), for j = 0, 1, ..., i, i+1.
Find the stationary distribution.
