ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Thu, 13 Jun 2013 10:24:40 +0200ergodic markov chain - steady statehttps://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/Hello every one
Is there anyway to define the ergodic markov chain or even the steady-state probability and select one state based on the given probability for each iteration??
Best
Aissan Tue, 11 Jun 2013 22:57:59 +0200https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/Comment by Bétréma for <p>Hello every one </p>
<p>Is there anyway to define the ergodic markov chain or even the steady-state probability and select one state based on the given probability for each iteration??</p>
<p>Best
Aissan </p>
https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?comment=17518#post-id-17518Question looks too general, ask Google about "coupling from the past".Wed, 12 Jun 2013 11:39:54 +0200https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?comment=17518#post-id-17518Answer by tmonteil for <p>Hello every one </p>
<p>Is there anyway to define the ergodic markov chain or even the steady-state probability and select one state based on the given probability for each iteration??</p>
<p>Best
Aissan </p>
https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?answer=15072#post-id-15072Iterating the matrix a lot is used to find the steady state, there is no iteration in picking a "random" element "in" the steady state, it is just about picking a random element in `[0,1]` and look where it falls according to the steady state, you won't lose much time there even for quite big matrices.
It should be also noticed that, since the matrix is positive, iterating will go quite fast to the dominating eigenvector (thanks to Perron Frobenius theorem there is no numerical precision issue since the corresponding line is attractive), this roughly corresponds to the ergodic hypothesis you mentioned. Hence, we are lucky to look for the dominating eigenvector. In our example:
sage: m^32
[0.259740259740260 0.285714285714286 0.454545454545454]
[0.259740259740260 0.285714285714286 0.454545454545454]
[0.259740259740260 0.285714285714286 0.454545454545455]
So 32 iterations are enough here to find the steady state.
Notice also that atlas can handle quite big matrices, if you want to take advantage of atlas, you also have to use a matrix with entries in `RDF` as well (for the same reason as for the `.left_eigenvectors()` method: some special software included in Sage only deal with the double precision floating point numbers that can be directly handled by the processor (not the one emulated by MPFR like in `RR`)). You can get those functionalities by simply doing:
sage: m = m.change_ring(RDF)
Thanks @ppurka for mentioning that `.left_eigenvectors()` method works with matrices in `RDF` i will update my answer accordingly.
Thu, 13 Jun 2013 10:24:40 +0200https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?answer=15072#post-id-15072Answer by tmonteil for <p>Hello every one </p>
<p>Is there anyway to define the ergodic markov chain or even the steady-state probability and select one state based on the given probability for each iteration??</p>
<p>Best
Aissan </p>
https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?answer=15063#post-id-15063A markov chain is just a stochastic matrix:
sage: m = matrix([[0.1, 0.2, 0.7], [0.5, 0.5, 0], [0.2, 0.2, 0.6]])
sage: m
[0.100000000000000 0.200000000000000 0.700000000000000]
[0.500000000000000 0.500000000000000 0.000000000000000]
[0.200000000000000 0.200000000000000 0.600000000000000]
As this one has positive entries, it is ergodic. Now, given a starting probability `P0`, you can iterate `m` on it:
sage: P0 = vector([0.2, 0.3, 0.5]) ; P0
(0.200000000000000, 0.300000000000000, 0.500000000000000)
sage: P = P0 * m^123 ; P
(0.259740259740260, 0.285714285714286, 0.454545454545454)
Now, if you want to sample according to this probability vector, you can make 3 boxes of the interval `[0,1]` whose length correspond to your probability vector, pick a uniformly random point in `[0,1]`, and check in which box it falls:
sage: def sample(P):
....: s = 0
....: a = random()
....: for box in range(len(P)):
....: s += P[box]
....: if a <= s:
....: return box
sage: sample(P)
1
sage: sample(P)
2
sage: sample(P)
0
sage: sample(P)
0
sage: sample(P)
1
sage: sample(P)
2
Looking for the steady-state probability, you can approach it by iterating the matrix `m` a lot:
sage: vector([1,0,0]) * m^1000
(0.259740259740259, 0.285714285714285, 0.454545454545454)
The convergence (quality and speed) is guaranteed by Perron Frobenius theorem.
If you are not sure how big should be the exponend to reach the steady-state, you can let Sage discover a fixed point by itself:
sage: def steady_state_iteration(m):
....: # just in case vector like (1,0,0,0) falls in the contracting space.
....: a = random()
....: V = vector([a, 1-a] + [0 for i in range(len(m[0])-2)])
....: while True:
....: W = V * m
....: W = W/sum(W)
....: if V == W:
....: return V
....: else:
....: V = W
sage: steady_state_iteration(m)
(0.25974025974, 0.285714285714, 0.454545454545)
You can also ask Sage for the dominating eigenvector. But you will notice that Sage does not accept to compute eigenvectors on the `Real Field` because no efficient and reliable was written there:
sage: m.left_eigenvectors()
NotImplementedError: eigenspaces cannot be computed reliably for inexact rings such as Real Field with 53 bits of precision,
consult numerical or symbolic matrix classes for other options
A first possibility is to work on an exact field like the rationals, and look for the (normalized) eigenvector whose eigenvalue is `1`:
sage: def steady_state_eigenvector(m):
....: for eig in m.left_eigenvectors():
....: if eig[0] == 1:
....: v = eig[1][0]
....: return v/sum(v)
sage: m = matrix([[1/10,2/10,7/10],[5/10,5/10,0],[2/10,2/10,6/10]]) ; m
[1/10 1/5 7/10]
[ 1/2 1/2 0]
[ 1/5 1/5 3/5]
sage: steady_state_eigenvector(m)
(20/77, 2/7, 5/11)
sage: steady_state_eigenvector(m).change_ring(RR)
(0.259740259740260, 0.285714285714286, 0.454545454545455)
So, you can check it is quite close to the numerical approach above.
Another possiblity (suggested by @ppurka in comments), is to work in the `Real Double Field`:
sage: m = m.change_ring(RDF)
sage: m.left_eigenvectors()
[(1.0, [(0.435503556309, 0.47905391194, 0.762131223541)], 1),
(-0.1, [(0.707106781187, 3.40584039133e-16, -0.707106781187)], 1),
(0.3, [(-0.348742916231, -0.464990554975, 0.813733471207)], 1)]
But, here you get,
sage: steady_state_eigenvector(m)
sage:
This is due to the fact that `1.0` given in the answer is not precisely equal to `1`:
sage: m.left_eigenvectors()[0][0] == 1
False
Hence, you have to look for approximate 1:
sage: def steady_state_eigenvector_approximate(m, tol=2^(-50)):
....: for eig in m.left_eigenvectors():
....: if (eig[0] - 1) <= tol:
....: v = eig[1][0]
....: return v/sum(v)
sage: steady_state_eigenvector_approximate(m)
(0.25974025974, 0.285714285714, 0.454545454545)
Wed, 12 Jun 2013 11:49:11 +0200https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?answer=15063#post-id-15063Comment by Bétréma for <div class="snippet"><p>A markov chain is just a stochastic matrix:</p>
<pre><code>sage: m = matrix([[0.1, 0.2, 0.7], [0.5, 0.5, 0], [0.2, 0.2, 0.6]])
sage: m
[0.100000000000000 0.200000000000000 0.700000000000000]
[0.500000000000000 0.500000000000000 0.000000000000000]
[0.200000000000000 0.200000000000000 0.600000000000000]
</code></pre>
<p>As this one has positive entries, it is ergodic. Now, given a starting probability <code>P0</code>, you can iterate <code>m</code> on it:</p>
<pre><code>sage: P0 = vector([0.2, 0.3, 0.5]) ; P0
(0.200000000000000, 0.300000000000000, 0.500000000000000)
sage: P = P0 * m^123 ; P
(0.259740259740260, 0.285714285714286, 0.454545454545454)
</code></pre>
<p>Now, if you want to sample according to this probability vector, you can make 3 boxes of the interval <code>[0,1]</code> whose length correspond to your probability vector, pick a uniformly random point in <code>[0,1]</code>, and check in which box it falls:</p>
<pre><code>sage: def sample(P):
....: s = 0
....: a = random()
....: for box in range(len(P)):
....: s += P[box]
....: if a <= s:
....: return box
sage: sample(P)
1
sage: sample(P)
2
sage: sample(P)
0
sage: sample(P)
0
sage: sample(P)
1
sage: sample(P)
2
</code></pre>
<p>Looking for the steady-state probability, you can approach it by iterating the matrix <code>m</code> a lot:</p>
<pre><code>sage: vector([1,0,0]) * m^1000
(0.259740259740259, 0.285714285714285, 0.454545454545454)
</code></pre>
<p>The convergence (quality and speed) is guaranteed by Perron Frobenius theorem.</p>
<p>If you are not sure how big should be the exponend to reach the steady-state, you can let Sage discover a fixed point by itself:</p>
<pre><code>sage: def steady_state_iteration(m):
....: # just in case vector like (1,0,0,0) falls in the contracting space.
....: a = random()
....: V = vector([a, 1-a] + [0 for i in range(len(m[0])-2)])
....: while True:
....: W = V * m
....: W = W/sum(W)
....: if V == W:
....: return V
....: else:
....: V = W
sage: steady_state_iteration(m)
(0.25974025974, 0.285714285714, 0.454545454545)
</code></pre>
<p>You can also ask Sage for the dominating eigenvector. But you will notice that Sage does not accept to compute eigenvectors on the <code>Real Field</code> because no efficient and reliable was written there:</p>
<pre><code>sage: m.left_eigenvectors()
NotImplementedError: eigenspaces cannot be computed reliably for inexact rings such as Real Field with 53 bits of precision,
consult numerical or symbolic matrix classes for other options
</code></pre>
<p>A first possibility is to work on an exact field like the rationals, and look for the (normalized) eigenvector whose eigenvalue is <code>1</code>:</p>
<pre><code>sage: def steady_state_eigenvector(m):
....: for eig in m.left_eigenvectors():
....: if eig[0] == 1:
....: v = eig[1][0]
....: return v/sum(v)
sage: m = matrix([[1/10,2/10,7/10],[5/10,5/10,0],[2/10,2/10,6/10]]) ; m
[1/10 1/5 7/10]
[ 1/2 1/2 0]
[ 1/5 1/5 3/5]
sage: steady_state_eigenvector(m)
(20/77, 2/7, 5/11)
sage: steady_state_eigenvector(m).change_ring(RR)
(0.259740259740260, 0.285714285714286, 0.454545454545455)
</code></pre>
<p>So, you can check it is quite close to the numerical approach above.</p>
<p>Another possiblity (suggested by <a href="/users/1186/ppurka/">@ppurka</a> in comments), is to work in the ...<span class="expander"> <a>(more)</a></span></p></div>https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?comment=17512#post-id-17512"Iterating a lot" is the obvious way for picking a "random" element "in" the steady state, problems arise with the specification of "a lot" :-) And in real life, matrices which describe Markov chains are rather large.Thu, 13 Jun 2013 09:12:51 +0200https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?comment=17512#post-id-17512Comment by tmonteil for <div class="snippet"><p>A markov chain is just a stochastic matrix:</p>
<pre><code>sage: m = matrix([[0.1, 0.2, 0.7], [0.5, 0.5, 0], [0.2, 0.2, 0.6]])
sage: m
[0.100000000000000 0.200000000000000 0.700000000000000]
[0.500000000000000 0.500000000000000 0.000000000000000]
[0.200000000000000 0.200000000000000 0.600000000000000]
</code></pre>
<p>As this one has positive entries, it is ergodic. Now, given a starting probability <code>P0</code>, you can iterate <code>m</code> on it:</p>
<pre><code>sage: P0 = vector([0.2, 0.3, 0.5]) ; P0
(0.200000000000000, 0.300000000000000, 0.500000000000000)
sage: P = P0 * m^123 ; P
(0.259740259740260, 0.285714285714286, 0.454545454545454)
</code></pre>
<p>Now, if you want to sample according to this probability vector, you can make 3 boxes of the interval <code>[0,1]</code> whose length correspond to your probability vector, pick a uniformly random point in <code>[0,1]</code>, and check in which box it falls:</p>
<pre><code>sage: def sample(P):
....: s = 0
....: a = random()
....: for box in range(len(P)):
....: s += P[box]
....: if a <= s:
....: return box
sage: sample(P)
1
sage: sample(P)
2
sage: sample(P)
0
sage: sample(P)
0
sage: sample(P)
1
sage: sample(P)
2
</code></pre>
<p>Looking for the steady-state probability, you can approach it by iterating the matrix <code>m</code> a lot:</p>
<pre><code>sage: vector([1,0,0]) * m^1000
(0.259740259740259, 0.285714285714285, 0.454545454545454)
</code></pre>
<p>The convergence (quality and speed) is guaranteed by Perron Frobenius theorem.</p>
<p>If you are not sure how big should be the exponend to reach the steady-state, you can let Sage discover a fixed point by itself:</p>
<pre><code>sage: def steady_state_iteration(m):
....: # just in case vector like (1,0,0,0) falls in the contracting space.
....: a = random()
....: V = vector([a, 1-a] + [0 for i in range(len(m[0])-2)])
....: while True:
....: W = V * m
....: W = W/sum(W)
....: if V == W:
....: return V
....: else:
....: V = W
sage: steady_state_iteration(m)
(0.25974025974, 0.285714285714, 0.454545454545)
</code></pre>
<p>You can also ask Sage for the dominating eigenvector. But you will notice that Sage does not accept to compute eigenvectors on the <code>Real Field</code> because no efficient and reliable was written there:</p>
<pre><code>sage: m.left_eigenvectors()
NotImplementedError: eigenspaces cannot be computed reliably for inexact rings such as Real Field with 53 bits of precision,
consult numerical or symbolic matrix classes for other options
</code></pre>
<p>A first possibility is to work on an exact field like the rationals, and look for the (normalized) eigenvector whose eigenvalue is <code>1</code>:</p>
<pre><code>sage: def steady_state_eigenvector(m):
....: for eig in m.left_eigenvectors():
....: if eig[0] == 1:
....: v = eig[1][0]
....: return v/sum(v)
sage: m = matrix([[1/10,2/10,7/10],[5/10,5/10,0],[2/10,2/10,6/10]]) ; m
[1/10 1/5 7/10]
[ 1/2 1/2 0]
[ 1/5 1/5 3/5]
sage: steady_state_eigenvector(m)
(20/77, 2/7, 5/11)
sage: steady_state_eigenvector(m).change_ring(RR)
(0.259740259740260, 0.285714285714286, 0.454545454545455)
</code></pre>
<p>So, you can check it is quite close to the numerical approach above.</p>
<p>Another possiblity (suggested by <a href="/users/1186/ppurka/">@ppurka</a> in comments), is to work in the ...<span class="expander"> <a>(more)</a></span></p></div>https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?comment=17510#post-id-17510No space enough for this comment box, see my comment below.Thu, 13 Jun 2013 10:15:14 +0200https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?comment=17510#post-id-17510Comment by ppurka for <div class="snippet"><p>A markov chain is just a stochastic matrix:</p>
<pre><code>sage: m = matrix([[0.1, 0.2, 0.7], [0.5, 0.5, 0], [0.2, 0.2, 0.6]])
sage: m
[0.100000000000000 0.200000000000000 0.700000000000000]
[0.500000000000000 0.500000000000000 0.000000000000000]
[0.200000000000000 0.200000000000000 0.600000000000000]
</code></pre>
<p>As this one has positive entries, it is ergodic. Now, given a starting probability <code>P0</code>, you can iterate <code>m</code> on it:</p>
<pre><code>sage: P0 = vector([0.2, 0.3, 0.5]) ; P0
(0.200000000000000, 0.300000000000000, 0.500000000000000)
sage: P = P0 * m^123 ; P
(0.259740259740260, 0.285714285714286, 0.454545454545454)
</code></pre>
<p>Now, if you want to sample according to this probability vector, you can make 3 boxes of the interval <code>[0,1]</code> whose length correspond to your probability vector, pick a uniformly random point in <code>[0,1]</code>, and check in which box it falls:</p>
<pre><code>sage: def sample(P):
....: s = 0
....: a = random()
....: for box in range(len(P)):
....: s += P[box]
....: if a <= s:
....: return box
sage: sample(P)
1
sage: sample(P)
2
sage: sample(P)
0
sage: sample(P)
0
sage: sample(P)
1
sage: sample(P)
2
</code></pre>
<p>Looking for the steady-state probability, you can approach it by iterating the matrix <code>m</code> a lot:</p>
<pre><code>sage: vector([1,0,0]) * m^1000
(0.259740259740259, 0.285714285714285, 0.454545454545454)
</code></pre>
<p>The convergence (quality and speed) is guaranteed by Perron Frobenius theorem.</p>
<p>If you are not sure how big should be the exponend to reach the steady-state, you can let Sage discover a fixed point by itself:</p>
<pre><code>sage: def steady_state_iteration(m):
....: # just in case vector like (1,0,0,0) falls in the contracting space.
....: a = random()
....: V = vector([a, 1-a] + [0 for i in range(len(m[0])-2)])
....: while True:
....: W = V * m
....: W = W/sum(W)
....: if V == W:
....: return V
....: else:
....: V = W
sage: steady_state_iteration(m)
(0.25974025974, 0.285714285714, 0.454545454545)
</code></pre>
<p>You can also ask Sage for the dominating eigenvector. But you will notice that Sage does not accept to compute eigenvectors on the <code>Real Field</code> because no efficient and reliable was written there:</p>
<pre><code>sage: m.left_eigenvectors()
NotImplementedError: eigenspaces cannot be computed reliably for inexact rings such as Real Field with 53 bits of precision,
consult numerical or symbolic matrix classes for other options
</code></pre>
<p>A first possibility is to work on an exact field like the rationals, and look for the (normalized) eigenvector whose eigenvalue is <code>1</code>:</p>
<pre><code>sage: def steady_state_eigenvector(m):
....: for eig in m.left_eigenvectors():
....: if eig[0] == 1:
....: v = eig[1][0]
....: return v/sum(v)
sage: m = matrix([[1/10,2/10,7/10],[5/10,5/10,0],[2/10,2/10,6/10]]) ; m
[1/10 1/5 7/10]
[ 1/2 1/2 0]
[ 1/5 1/5 3/5]
sage: steady_state_eigenvector(m)
(20/77, 2/7, 5/11)
sage: steady_state_eigenvector(m).change_ring(RR)
(0.259740259740260, 0.285714285714286, 0.454545454545455)
</code></pre>
<p>So, you can check it is quite close to the numerical approach above.</p>
<p>Another possiblity (suggested by <a href="/users/1186/ppurka/">@ppurka</a> in comments), is to work in the ...<span class="expander"> <a>(more)</a></span></p></div>https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?comment=17511#post-id-17511It is not that sage does not *like* to compute eigenvalues in floating point fields; it is just that it can not guarantee the required precision in RR and be fast at the same time. You can specify everything to belong to RDF instead and sage will happily compute the eigenvectors for you.
sage: matrix(RDF, [[1.0, 1.5],[2, 4]]).left_eigenvectors()
[(0.208712152522, [(-0.929866873266, 0.367896178294)], 1),
(4.79128784748, [(-0.466583903687, -0.884476941938)], 1)]
Also, ticket 13660 is relevant in this context.Thu, 13 Jun 2013 09:35:22 +0200https://ask.sagemath.org/question/10222/ergodic-markov-chain-steady-state/?comment=17511#post-id-17511