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I have a full set of sequences (432 observations to be precise) of 4 states $A-D$: eg

$$Y=\left(\begin{array}{c c c c c c c} A& C& D&D & B & A &C\\ B& A& A&C & A&- &-\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ B& C& A&D & A & B & A\\ \end{array}\right)$$

EDIT: The observation sequences are of unequal lengths! Does this change anything?

Is there a way of calculating the transition matrix $$P_{ij}(Y_{t}=j|Y_{t-1}=i)$$ in Matlab or R or similar? I think the HMM package might help. Any thoughts?

eg: Estimating Markov chain probabilities

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    $\begingroup$ You have $4$ states: $S=\{1:=A,2:=B,3:=C,4:=D\}$. Let $n_{ij}$ be the number of times the chain made a transition from state $i$ to state $j$, for $ij,=1,2,3,4$. Compute the $n_{ij}$'s from your sample and estimate the transition matrix $(p_{ij})$ by maximum likelihood using the estimates $\hat{p}_{ij}=n_{ij}/\sum_{j=1}^4 n_{ij}$. $\endgroup$
    – Zen
    Commented Sep 11, 2012 at 16:29
  • $\begingroup$ These notes derive the MLE estimates: stat.cmu.edu/~cshalizi/462/lectures/06/markov-mle.pdf $\endgroup$
    – Zen
    Commented Sep 11, 2012 at 16:30
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    $\begingroup$ Similar question:stats.stackexchange.com/questions/26722/… $\endgroup$
    – B_Miner
    Commented Sep 11, 2012 at 20:18
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    $\begingroup$ @B_Miner could you write your code in pseudo-code form for me? Or explain it in lay terms... However I see it works in my R console. $\endgroup$
    – HCAI
    Commented Sep 11, 2012 at 21:41
  • $\begingroup$ I have a question: I understand your implementation and it lokks fine to me, but i was wondering why can't i simply use the Matlab hmmestimate function to compute the T matrix? Something like: states=[1,2,3,4] [T,E]= hmmestimate ( x, states); where T is the transition matrix i'm interested in. I'm new to Markov chains and HMM so I'd like to understand the difference between the two implementations (if there is any). $\endgroup$
    – Any
    Commented Nov 20, 2013 at 11:53

3 Answers 3

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Please, check the comments above. Here is a quick implementation in R.

x <- c(1,2,1,1,3,4,4,1,2,4,1,4,3,4,4,4,3,1,3,2,3,3,3,4,2,2,3)
p <- matrix(nrow = 4, ncol = 4, 0)
for (t in 1:(length(x) - 1)) p[x[t], x[t + 1]] <- p[x[t], x[t + 1]] + 1
for (i in 1:4) p[i, ] <- p[i, ] / sum(p[i, ])

Results:

> p
          [,1]      [,2]      [,3]      [,4]
[1,] 0.1666667 0.3333333 0.3333333 0.1666667
[2,] 0.2000000 0.2000000 0.4000000 0.2000000
[3,] 0.1428571 0.1428571 0.2857143 0.4285714
[4,] 0.2500000 0.1250000 0.2500000 0.3750000

A (probably dumb) implementation in MATLAB (which I have never used, so I don't know if this is going to work. I've just googled "declare vector matrix MATLAB" to get the syntax):

x = [ 1, 2, 1, 1, 3, 4, 4, 1, 2, 4, 1, 4, 3, 4, 4, 4, 3, 1, 3, 2, 3, 3, 3, 4, 2, 2, 3 ]
n = length(x) - 1
p = zeros(4,4)
for t = 1:n
  p(x(t), x(t + 1)) = p(x(t), x(t + 1)) + 1
end
for i = 1:4
  p(i, :) = p(i, :) / sum(p(i, :))
end
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  • $\begingroup$ Looks great! I'm not sure what the 3rd line does in your code though (mainly because I'm familiar with Matlab). Any chance you could write it in matlab or pseudo-code? I'd be much obliged. $\endgroup$
    – HCAI
    Commented Sep 11, 2012 at 17:22
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    $\begingroup$ The third line does this: the chain values are $x_1,\dots,x_n$. For $t=1,\dots,n-1$, increment $p_{x_t,x_{t+1}}$. $\endgroup$
    – Zen
    Commented Sep 11, 2012 at 19:32
  • $\begingroup$ The fourth line normalizes each line of the matrix $(p_{ij})$. $\endgroup$
    – Zen
    Commented Sep 11, 2012 at 19:34
  • $\begingroup$ Bare with my slowness here. I do appreciate the MATLAB code translation although I still can't see what it's attempting to do in your first for loop. The 3rd line from the original code is counting the number of times $x$ goes from state $x_i$ to state $x_j$? If you could say it in words I'd appreciate that a lot. Cheers $\endgroup$
    – HCAI
    Commented Sep 11, 2012 at 20:35
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    $\begingroup$ No, $x$ is just one row. Don't concatenate because you will introduce "false" transitions: last state of one line $\to$ first state of the next line. You have to change the code to loop through the lines of your matrix and count the transitions. At the end, normalize each line of the transition matrix. $\endgroup$
    – Zen
    Commented Sep 15, 2012 at 23:57
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Here is my implementation in R

x <- c(1,2,1,1,3,4,4,1,2,4,1,4,3,4,4,4,3,1,3,2,3,3,3,4,2,2,3)
xChar<-as.character(x)
library(markovchain)
mcX<-markovchainFit(xChar)$estimate
mcX
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    $\begingroup$ user32041's request (posted as an edit instead of a comment since he/she lacks reputation): How can I coerce the transitionMatrix of the markovchainFit result to a data.frame? $\endgroup$
    – chl
    Commented Oct 29, 2013 at 14:33
  • $\begingroup$ You can convert to $data.frame$ using $as(mcX,"data.frame")$ $\endgroup$
    – Giorgio Spedicato
    Commented Nov 14, 2013 at 23:23
  • $\begingroup$ @GiorgioSpedicato can you comment on how to deal with sequences of unequal lengths (I cannot concatenate) please in your package? $\endgroup$
    – HCAI
    Commented Aug 12, 2019 at 18:52
  • $\begingroup$ @HCAI, please see the current vignette page 35-36 $\endgroup$
    – Giorgio Spedicato
    Commented Aug 12, 2019 at 20:54
  • $\begingroup$ @GiorgioSpedicato thank you for the reference cran.r-project.org/web/packages/markovchain/vignettes/…. I still have n transition matrices, one for each sequence. What I’m after is one general one that takes into account all the sequence observations. Is there something I’m missing? $\endgroup$
    – HCAI
    Commented Aug 13, 2019 at 3:37
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Here is a way to do it in Matlab:

x = [1,2,1,1,3,4,4,1,2,4,1,4,3,4,4,4,3,1,3,2,3,3,3,4,2,2,3];
counts_mat = full(sparse(x(1:end-1),x(2:end),1));
trans_mat = bsxfun(@rdivide,counts_mat,sum(counts_mat,2))

Acknowledgement owed to SomptingGuy: http://www.eng-tips.com/viewthread.cfm?qid=236532

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