1

I am currently having some problems on the following question:

Given is the function $f(x)$:

$f(x) = 0,1,2$ with probability $\frac{1}{3}$ for each.

I have to give the state space, transition probability matrix and explain why independent successive draws from $f(x)$, $X_1,X_2,\dots$, is a time-homogeneous Markov Chain.

This is really unclear to me since it seems to me that it is not a Markov Chain as the next state is independent from the previous state.

Also, would the transition matrix be a $3$ x $3$ matrix with each entry $\frac{1}{3}$?

Thank you in forward!

Nedellyzer
  • 1,174
  • I think you are correct. You can realize the $ij$th entry as the probability of transitioning from the $j$th state to the $i$th state. There is no correlation so I think each element should be $\frac{1}{3}$. – Cameron Williams Jan 24 '15 at 19:32
  • @CameronWilliams Thanks for your reply. But would it be a Markov Chain in this case since I thought the next state is dependent on the previous state? – Nedellyzer Jan 24 '15 at 19:34
  • I think the Markov condition is actually that the probabilities only rely on the immediately preceding time step, not time steps prior to that. – Cameron Williams Jan 24 '15 at 19:37
  • @CameronWilliams Yes exactly, but there is no dependence on the immediately preceding time step, am I right? – Nedellyzer Jan 24 '15 at 19:38
  • There isn't, but I don't think it matters. Would you say that $f(x) = 1$ is a function of $x$ even though it only has constant values? – Cameron Williams Jan 24 '15 at 19:40
  • @CameronWilliams All right, I see where you are going. I was only assuming that a condition for a Markov Chain would be that state $i$ is correlated/dependent on state $i-1$? – Nedellyzer Jan 24 '15 at 19:46
  • Yes. $p(x,t)$ need not be dependent upon $p(x,t-1)$, $p(x,t-2)$, etc. but if it is, $p(x,t)$ can only be dependent upon $p(x,t-1)$. – Cameron Williams Jan 24 '15 at 19:48

0 Answers0