0

If I have a variable X, that follows a Markov Chain with a transition density $\rho(X)$ does a function of that variable f(X) have the same density or is there a one to one mapping to the density of f(X) from $\rho(X)$?

Drew
  • 109

2 Answers2

0

Let $X$ be a Markov process. First of all, $f(X)$ may not be a Markov process: think of $X$ being a skewed random walk on $\Bbb Z$ which moves to the right with probability $\frac23$. Then if $f(x) = x^2$ and you know that $X^2_n = 1$ you can't really say what's the probability that $X^2_{n+1} = 2$: for that you need to know the sign of $X_n$ which you can't deduce from $X^2_n$ alone.

You can always show that $f(X)$ is Markov whenever $f$ is injection, if $f$ is not an injection, then $f(X)$ will be Markov just because the transition probabilities of $X$ are "same" on the level sets of $f(X)$. There is a whole area of lumping/bisimulation where they study how to find such non-injective $f$ for a given Markov process. Anyways, when $f$ is injective you shall be able to derive the transition matrix for $f(X)$ from that of $X$.

Edit: Due to the issue with a biased random walk mentioned by @Did, consider another example. Let's $X_{n+1} = 2X_n$ and $X_0$ is uniformly distributed on $\{-1,1\}$. Let $f(x) = x1_{|x|\neq 2} + |x|1_{|x| = 2}$. Then we obtain that $\mathsf P(f_2 = 4|f_0,f_1) = 1_{f_0 = 1}$ whereas $\mathsf P(f_2 = 4|f_1) = \frac12$.

SBF
  • 36,568
  • The example of a biased random walk on the integers is not so inspiring since in fact its absolute value is Markov, if one starts from $0$ or if one looks at the path after it first hits $0$. The transitions of this Markov chain are $0\to1$ with probability $1$, $n\to n+1$ with probability $(p^{n+1}+q^{n+1})/(p^n+q^n)$ for every $n\geqslant1$, and $n\to n-1$ with the complementary probability. For the Brownian analogue of this result, see http://math.stackexchange.com/q/1004864/. – Did Nov 26 '14 at 21:49
  • @Did: I see, my example only says that filtration of $X$ contains more information about $f(X)$ than the natural filtration of $f(X)$. Is the new example correct? Also, regarding the square of a biased random walk - does Markovianity depend on the initial distribution? I'm not sure how to show that conditional distributions are the same. – SBF Nov 27 '14 at 08:41
0

If $f$ is one-to-one, $P(f(X_{n+1}) = f(t) | f(X_n) = f(s)) = P(X_{n+1} = t | X_n = s)$. So the transition matrix for $f(X_n)$ is obtained from that for $X_n$ by relabeling the rows and columns.

Robert Israel
  • 470,583
  • I am trying to prove the same result using the definition of Markov process with $\sigma$-algebra, but I am not sure how to use $f$ is an injection. Does it implies $P(f(X_{n+1}) \in B | f(X_n) ) = P(X_{n+1} \in f^{-1}(B) | X_n)$? I don't know how to show that. – John Feb 12 '16 at 13:59
  • 1
    $f(X) \in B$ is always the same as $X \in f^{-1}(B)$. If $f$ is an injection and $f$ and $f^{-1}$ are Borel functions, the $\sigma$-algebras generated by $f(X)$ and $X$ are the same. – Robert Israel Feb 12 '16 at 16:23
  • Thanks for your response. But my goal is to show if $X_n$ is Markov, and $f$ is injective and Borel measurable, then $f(X_n)$ is Markov. Does these assumptions already imply $f^{-1}$ is Borel? In my opinion, it's not since we need to prove $f(B)$ is Borel for any Borel set $B$, which is different from $f$ is Borel measurable. – John Feb 12 '16 at 17:08
  • 1
    There is a theorem of Sierpinski that a function from $\mathbb R$ to $\mathbb R$ is Borel measurable iff its graph is a Borel set. So if $f$ is injective and Borel measurable, $f^{-1}$ is also Borel measurable. – Robert Israel Feb 12 '16 at 18:21
  • So the theorem also works if domain of $f$ is a subset of $\mathbb{R}$? Because when I read the linked book, it seems that it always assume $f$ is also onto. – John Feb 13 '16 at 00:59