I have given a Markov Chain $X_n$ with the state space $\{0,1,2\}$ and the transition Matrix $$P= \begin{Bmatrix} 0.3 & 0.2 & 0.5 \\ 0.5 & 0 & 0.5 \\ 0.2 & 0.1 & 0.7 \end{Bmatrix} $$

Given a function $f(x)=x^3$ for $x\in\{ 0,1,2 \}$. Is $f(X_n)$ a Markov Chain?

I don't know what to do here. I looked in my lectures but can't find anything on what a function does to a Markov Chain. I know that the state space changes to $\{0,1,8\}$ but what does the function do with the transition probabilities? How does the transition matrix change? Is $f(X_n)$ even a Markov chain? And if not how do you see this?

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    $\begingroup$ Hint: The map $x\mapsto x^3$ on $\{0,1,2\}$ is injective. $\endgroup$ – Math1000 Feb 11 '16 at 20:34
  • $\begingroup$ So what does that mean? I still don't know how the transition matrix changes. $\endgroup$ – Lawg Feb 11 '16 at 20:47
  • $\begingroup$ It doesn't change the transition probabilities. See @grand_chat's answer. $\endgroup$ – Math1000 Feb 12 '16 at 1:31

Since $f$ is one-to-one, the new sequence $f(X_n)$ is still a Markov chain. To prove this, argue that the transition matrix is the same as before, only the rows and columns are labelled with the new state space $f(0), f(1), f(2)$ instead of $0, 1, 2$.

  • $\begingroup$ ok I think I understand why an one-on-one function of the Markov chain is still a markov chain. But why does the transition matrix stay the same? The problem is that I can't see a connection between $f(X_n)$ and the transition probabilities.... $\endgroup$ – Lawg Feb 11 '16 at 21:15
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    $\begingroup$ For example: $P(f(X_2)=f(x_2) \mid f(X_1)=f(x_1)) = P(X_2=x_2\mid X_1=x_1)$ for any $x_1$ and $x_2$. The reason is $f$ is one-to-one, so the event $f(X_2)=f(x_2)$ is the same as the event $X_2=x_2$, and so on. $\endgroup$ – grand_chat Feb 11 '16 at 21:19
  • $\begingroup$ thanks much! One last question: if it was for example $f(x)=x^2$ with the state space $\{-1,0,1\}$ and the same transition matrix. Is then $f(X_n)$ a markov chain? The new state space is then $\{0,1\}$ and $f$ is not one-on-one. $\endgroup$ – Lawg Feb 11 '16 at 21:48
  • $\begingroup$ The answer is maybe yes, maybe no. You have to calculate the transition probabilities to find out. Call the new sequence $Y_1, Y_2, \ldots$. To be Markov the transition probs $P(Y_2=y_2\mid Y_1=y_1)$ have to be independent of your starting distribution $P(Y_1=y_1)$, for all possible values of $y_1$ and $y_2$. This calculation will be tedious. For example calculate $P(Y_2=1, Y_1=1) = P(Y_2=1, X_1=1) + P(Y_2=1, X_1=-1)$ = $P(Y_2=1\mid X_1=1)P(X_1=1) + P(Y_2=1\mid X_1=-1)P(X_1=-1)$ = $.9P(X_1=1) + .8P(X_1=-1)$. Divide this by $P(Y_1=1)=P(X_1=1) + P(X_1=-1)$ to obtain $P(Y_2=1\mid Y_1=1)$. $\endgroup$ – grand_chat Feb 11 '16 at 23:16

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