Timeline for Compute the stationary distribution of a large transition matrix
Current License: CC BY-SA 4.0
10 events
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| Nov 14, 2020 at 18:07 | comment | added | LeafGlowPath | @ChrisK Yes. This is very fast. Thanks. | |
| Nov 14, 2020 at 17:40 | comment | added | Chris K |
@Roman Upon closer reading of the problem, it's clear that it's a discrete-time Markov chain, so the solution is even easier (and doesn't require IdentityMatrix at all).
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| Nov 14, 2020 at 17:39 | history | edited | Chris K | CC BY-SA 4.0 |
even simpler in discrete-time
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| Nov 14, 2020 at 17:35 | comment | added | Roman | Agreed, you're right that if the matrix is constructed correctly, it will have the right spectral property. | |
| Nov 14, 2020 at 17:32 | comment | added | Chris K |
I should add, I'm a bit unclear on OP's actual problem, whether it is a continuous- or discrete-time process and the role of the -IdentityMatrix[n] part here.
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| Nov 14, 2020 at 17:17 | comment | added | Chris K | @Roman True in general, but in this case I don't think it's luck, because all eigenvalues should be nonpositive except the one associated with the stationary distribution, which should be zero (conservation of probability). See cims.nyu.edu/~holmes/teaching/asa19/handout_Lecture4_2019.pdf for that last point. At least I hope this is true, because we're using this technique in a paper now :) | |
| Nov 14, 2020 at 16:46 | comment | added | Roman |
In this case it would be advantageous to point the Arnoldi algorithm towards the "right" eigenvalue by using a Shift operation: see this answer. In the present example you got lucky to catch the right eigenvalue.
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| Nov 14, 2020 at 16:22 | history | edited | Chris K | CC BY-SA 4.0 |
added 4 characters in body
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| Nov 14, 2020 at 15:29 | history | edited | Chris K | CC BY-SA 4.0 |
added explanation
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| Nov 14, 2020 at 15:21 | history | answered | Chris K | CC BY-SA 4.0 |