# Conditional transition matrix (consistent estimates)

Is there a model/technique that is able to estimate transition matrix (which would be consistent, i.e. sums of their rows would be always 1) conditional on some continuous variable X?

Let's say I have a system that can acquire one of 3 possible states (A,B,C). I have some observations and I could apply standard Markov chains. This way I can get something we could call "unconditional" transition matrix.

However, the transition probabilities (probably) depend on a continuous variable $X\in(0,1)$. The higher X is, the higher is the chance of A going to B and B going to C (C is absorption state).

These model the probability of being in a particular state on various distributions, e.g. on a continuous variable X∈(0,1). R has a nice package named MSwM for this.

There is also the concept of Hidden Markov Models where the conditional transition probabilities are quantified. However, as the name suggests, these only apply where you have 'latent' (i.e. hidden) states - which does not seem to be your case.

This is a time-inhomogeneous Markov Jump Chain

Transition matrix,$$P_{ij}(t)=\pmatrix{1-p(t)&p(t)&0\\0&1-p(t)&p(t)\\0&0&1}$$

As $X \in (0,1)$ , so for every $X(t)$ you'll get $p(t)$ so that you can make up.

• This answer has many problems. The question asks how to estimate this matrix (presumably using data). Your formulation doesn't meet the constraints on the transition probabilities, either. And your "$X$" does not appear to mean the same thing as the "$X$" in the question.
– whuber
Apr 29, 2015 at 16:23
• @whuber As he said X belongs (0,1) , so for every X(t) he'll get p(t) so that he can make up. Apr 29, 2015 at 19:01
• That's not anywhere close to what your notation says!
– whuber
Apr 29, 2015 at 19:03
• @whuber thanks! But I am wonder what my notations said? Apr 29, 2015 at 19:09
• "$p(t)\sim X(t)$", at least in a statistical context, would suggest that $X(t)$ is a family of distributions indexed by $t$, $p(t)$ is a family of random variables indexed by $t$, and that each $p(t)$ follows the distribution $X(t)$.
– whuber
Apr 29, 2015 at 19:12