6
$\begingroup$

A Markov Matrix is a square matrix,which have two features:

  • All elements great than or equal to $0$(But I hope all element great than or equal to $0.1$)
  • All the columns add up to $1$

I have a custom function for this

MarkovMatrix[dim_] := 
 Module[{m}, Label[start]; 
  m = Transpose[
    Append[#, 1 - Total[#]] & /@ 
     RandomReal[1, {dim, dim - 1}, WorkingPrecision -> 2]]; 
  If[AllTrue[m, # > .1 &, 2], m, Goto[start]]; m]

Usage:

For example to generate a 4*4 matrix

MatrixForm[m = MarkovMatrix[4]]

But my MarkovMatrix is low efficiency function.Are there any better method can do this?

$\endgroup$
5
  • $\begingroup$ Surely you mean the columns add up to $1$ $\endgroup$
    – Feyre
    Commented Jun 27, 2016 at 18:18
  • 2
    $\begingroup$ Transpose[Normalize[#, Total] & /@ RandomReal[1, {4, 4}]]? Anyway: the more common term is stochastic matrix. $\endgroup$ Commented Jun 27, 2016 at 18:18
  • $\begingroup$ @Feyre OMG.I make a typo.Thanks for your reminder. $\endgroup$
    – yode
    Commented Jun 27, 2016 at 18:22
  • $\begingroup$ @J.M. I have to say this is a beautiful solution.It deserve a answer but a comment.Another extra my request can you make all element great than or equal to $0.1$ with same elegant method? (Thanks for your term. :) $\endgroup$
    – yode
    Commented Jun 27, 2016 at 18:29
  • 2
    $\begingroup$ relevant: mathematica.stackexchange.com/questions/69707/… and mathematica.stackexchange.com/q/33652/2079. And for heavens sake never use Goto. $\endgroup$
    – george2079
    Commented Jun 27, 2016 at 19:59

3 Answers 3

10
$\begingroup$

Here's something even more compact than my proposal in the comments:

Standardize[RandomReal[1, {4, 4}], 0 &, Total]

If you must have a left stochastic matrix where all the entries should be greater than a set value, you can do rejection sampling: keep generating a matrix as long as the smallest value is smaller than the cutoff:

While[Min[sm = Standardize[RandomReal[1, {4, 4}], 0 &, Total]] < 0.1]; sm

If a doubly stochastic matrix is desired (that is, all columns and all rows sum to unity), some more trickery is necessary:

While[Min[dsm = FixedPoint[Standardize[Transpose[Standardize[#, 0 &, Total]],
                                       0 &, Total] &, RandomReal[1, {4, 4}],
                           SameTest -> (Norm[#1 - Transpose[#2], "Frobenius"] <
                                        1.*^-12 &)]] < 0.1]; dsm

I make no guarantees on the distribution followed by the matrices generated by either method.

$\endgroup$
0
8
$\begingroup$

Assuming you want uniformly distributed n-dimensional probability vectors with a minimum value, I think you can use:

MarkovMatrix[n_, min_:0] := If[min n<1,
    Transpose @ RandomPoint[Simplex[IdentityMatrix[n](1-min n)], n] + min,
    $Failed
]

For example:

MarkovMatrix[4, .1] // TeXForm

$\left( \begin{array}{cccc} 0.378616 & 0.267013 & 0.416824 & 0.142604 \\ 0.14229 & 0.305494 & 0.177654 & 0.203273 \\ 0.231628 & 0.175853 & 0.154734 & 0.178135 \\ 0.247466 & 0.251641 & 0.250788 & 0.475988 \\ \end{array} \right)$

$\endgroup$
2
  • $\begingroup$ For earlier versions that do not have RandomPoint[], here is some equivalent code: MarkovMatrix[n_, min_: 0] := If[min n < 1, Transpose[((1 - min n) Append[#, 1 - Total[#]]) & /@ RandomVariate[DirichletDistribution[ConstantArray[1, n]], n]] + min, $Failed] $\endgroup$ Commented Jul 26, 2017 at 14:09
  • $\begingroup$ Just discovered the //TeXForm command. I cried, for real. $\endgroup$
    – R.W
    Commented May 6, 2020 at 0:36
4
$\begingroup$

Here is my idea

make[n_] := ConstantArray[0.1, {n, n}] + (
                 (1 - 0.1 n) #/Total[#, {2}] &[RandomReal[{0, 1}, {n, n}]])

Because of the minimum you specified for each entry, this works for $n<11$ only. Did you consider letting the minimum value dependent on $n$ in a decreasing manner?

$\endgroup$
1
  • $\begingroup$ Thanks for your solution.And I have no that thinking,I just don't hope the element too tiny,which will make some troubles for my processing in following. $\endgroup$
    – yode
    Commented Jun 27, 2016 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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