1
$\begingroup$

A dice is thrown an infinite number of times. Which of the following procsses are Markov chains or not? Justify your answer. For those processes that are Markov chains give the transition probabilities. For $n\in\mathbb{N}$ consider

(1) $U_n$ to be the maximal number shown up to time $n$.

(2) $V_n$ to be the number of times a 6 appears within the first $n$ throws.

(3) $X_n$ to be the time which has passed since the last appearance of a $1$.

(4) $Y_n$ to be the time which will pass until the next $4$ appears.


A complete edit of my ideas

In my opinion all are Markov processes.

(a) is a Markov provess with state space $E=\left\{1,2,3,4,5,6\right\}$, because the Markov property is fullfilled:

In order to predict $U_{n+1}$ I only need to know the maximal number shown up to time $n$, i.e. $U_n$, because then I know that $U_{n+1}$ is either one of the numbers that are $> U_{n}$ or it is equal to $U_n$. I do not have to know $U_1,...,U_{n-1}$.

The transition matrix is $$ P=\begin{pmatrix}1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6\\0 & 1/3 & 1/6 & 1/6 & 1/6 & 1/6\\0 & 0 & 1/2 & 1/6 & 1/6 & 1/6\\ 0 & 0 & 0 & 2/3 & 1/6 & 1/6\\ 0 & 0 & 0 & 0 & 5/6 & 1/6\\ 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix} $$

(b) is a Markov chain with state space $E=\mathbb{N}\cup\left\{0\right\}$; the Markov property is fullfilled:

In order to predict $V_{n+1}$ I only have to know the number of times a 6 appears within the first $n$ throws, i.e. I have to know $V_n$, because then I know that $V_{n+1}$ is either $V_{n}+1$ or remains $V_n$.

Here the transition probabilities are $$ p_{ij}=\begin{cases}1/6, & j=i+1\\5/6, & i=j\end{cases}. $$

(c) is a Markov chain with state space $E=\mathbb{N}\cup\left\{0\right\}$. Again the Markov property is fullfilled:

If I want to predict $X_{n+1}$ (which I do understand as the time at throw number $n+1$ that has passed since the last 1 appeared), I only need to know the time that has passed at throw number $n$ since the last 1 appeared. Then I can predict that $V_{n+1}=0$ (if I throw a 1 at throw number $n+1$) or $V_{n+1}=V_n+1$.

So to my opinion the transition probabilities are given by $$ p_{ij}=\begin{cases}1/6, & j=0\\5/6, & j=i+1\end{cases}. $$

(d) Here I am not that sure. But I think here we have a Markov chain, too. Because I think here we do not need the past throws at all, because the time to the next throw of a $4$ does not depend at all from the past throws. So it makes no difference if we conditional on the whole past or only on $Y_n$ in order to predict $Y_{n+1}$. So the Markov property is fullfilled.

To my opinion the transition probabilities are $$ p_{ij}=\begin{cases}1/6, & j=1\\5/6, & j=i+1\end{cases}. $$

Would be really nice to hear from you, if I am right or not.

Greetings

$\endgroup$
  • $\begingroup$ Aren't your results and drawing past the horizontal line about the third process, and not the fourth? $\endgroup$ – zarathustra Oct 24 '14 at 16:42
  • $\begingroup$ If (a,b,c,d)=(1,2,3,4), the current answers are correct. The justification of answer d/4 is bogus, though, a full answer requires to make explicit the filtration for which Y is Markov. Let me suggest to think again about the transition matrix in case d/4. $\endgroup$ – Did Oct 31 '14 at 8:57
2
$\begingroup$

You say "because the future depends on the past," but this is not the correct characterization of a Markov Chain: for a Markov chain if you condition on the past, it only depends on the last value. So for $(2)$, $V_n$ conditioned on $V_{n-1},\cdots,V_1$, clearly depends only on $V_{n-1}$. Afterall, the number of 6's that occur in n throws satisfies $V_n=V_{n-1}+T_n$, where $T_n=1$ if the n'th throw is a 6 and 0 otherwise. Since dice throws are independent, $T_n$ is independent of $V_{n-1},\cdots,V_1$, making $P(V_n|V_{n-1},\cdots,V_1)=P(V_n|V_{n-1})$.

The same goes for (3). Since your throws are independent, this is what's called a renewal process: every time you throw a 1, it resets and is indepedent of everything prior. If i tell you that 20 minutes have passed since the last 1, then you don't need to know what happened 19 minutes ago, because the next throw is independent of the entire past, so if you want the new time passed since the last throw, you need only the last time update. In math language, the time $\tau_n$ since the last 1 was thrown has the following property: $\tau_n-\tau_{n-1}$ is independent of all $\tau_{i}-\tau_{i-1}$ for $i=1,2,\cdots,n-1$.

$\endgroup$
  • $\begingroup$ Thank you, you are right. I did not use the Markov property correctly. Now I see that (2) is a Markov chain, too. And then (3) is a Markov chain, too. If I condition on the past I only need the last value. Right? But am I right, that (3) and (4) do have the same transition probabilites? $\endgroup$ – mathfemi Oct 24 '14 at 17:18
  • $\begingroup$ Ok, I edit my post. Pls have a look on it then. $\endgroup$ – mathfemi Oct 24 '14 at 17:29
  • $\begingroup$ I changed my opinion, I think (1) is a Markov chain, too. Because i only need the last maximal number shown up to time n, i.e. $U_n$, to predict $U_{n+1}$. Right? $\endgroup$ – mathfemi Oct 30 '14 at 19:16
  • $\begingroup$ Please have a look on my edited question/ ideas. They changed. $\endgroup$ – mathfemi Oct 30 '14 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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