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Let $I=[0,1]$ and $\alpha \in ( 0,1)$. Define the Tent map, $T: I \rightarrow I$ by

$T(x)= x/\alpha$ for $x \in [0,\alpha]$ and $(1-x)/(1-\alpha)$ for $x \in [\alpha,1]$

Find the measure theoretic entropy.

For the case $\alpha =1/2$, I can calculate the Entropy. However, for the general case I am not sure how to do it. Since as soon as I start calculating $T^{-i}P$ for a given partition $P$ the expressions become very complicated.

I am guessing that I should move to a representation in terms of shift dynamics but I am not sure how. Could anybody give me a hint?

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You work with lebesgue measure, I guess. Consider the generating partition $\{[0, \alpha),[\alpha,1)\}$, or (this is roughly equivalent) prove that your system is isomorphic to a bernoulli shift on two symbols. –  D. Thomine Jan 17 '12 at 10:36
    
Thank you for your reply. But I had already tried this. The problem that I had was that I couldn't construct a conjugacy between the two. Since the conjugacy in terms of a binary expantion for the case $\alpha=1/2$ didn't work anymore. Could I have another hint? –  Ivah Jan 17 '12 at 12:14
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1 Answer

Before answering your question, I'll recall what happens with the dyadic map.

The case $\alpha=1/2$

Let $x$ be in $[0,1)$. We can write a proper binary expansion for $x$; in other words, we have a surjection $\varphi: \{0,1\}^{\mathbb{N}} \to [0,1)$ such that:

$$\varphi ((x_k)_{k \in \mathbb{N}}) = \sum_{k=1}^{+ \infty} x_k 2^{-k}.$$

If we endow $\{0,1\}^{\mathbb{N}}$ with the product measure giving the same weight to $0$'s and $1$'s ("heads or tails") and $[0,1)$ with the Lebesgue measure, then this is an isomorphism of measure spaces. The Bernoulli shift $\sigma : (x_k) \mapsto (x_{k+1})$ is then conjugated with the dyadic map $T:x \mapsto 2x [1]$. The dyadic map is basically the same map as your for $\alpha = 1/2$.

Now, we want to do the reverse: we start from the dyadic map, and try to encode it somehow. If we put $A_0 = [0,1/2)$ and $A_1=[1/2,1)$, then we can write:

$$\varphi^{-1} (x) = (1_{A_1} (T^k (x)))_{k \in \mathbb{N}}.$$

In other words, we track the orbit of $x$ by looking, at each step, into which element of the partition $\{A_0, A_1\}$ the point $T^k (x)$ is. We can do an even more abstract coding, such as $x \sim (b_k)_{k \in \mathbb{N}}$ where each $b_k$ takes it value into $\{A_0, A_1\}$. This step is very convenient: we can still see the dyadic map as a Bernoulli shift on $\{A_0, A_1\}^{\mathbb{N}}$, but we don't need this kind of algebraic miracle which links directly to the binary expansion. This construction can be generalized to a larger number of symbols.

The general case

As I said in my comment, a nice partition is $\{A_0, A_1\}$, where $A_0 = [0,\alpha)$ and $A_1=[\alpha,1)$. Well, you can do pretty much what I said above : since this partition generates the usual $\sigma$-algebra on $[0,1)$, you can encode any point $x$ in $[0,1)$ by a sequence $(b_k)_{k \in \mathbb{N}}$, where $b_k = A_i$ if an only if $T^k (x) \in A_i$.

Now, the pertinent question is: what is the push-forward of the Lebesgue measure by this transformation? If we denote by $\mu$ this measure, then the systems $([0,1), \text{Leb}, T)$ and $(\{A_0, A_1\}^\mathbb{N}, \mu, \sigma)$, where $\sigma$ is the shift, are isomorphic.

Good news: this is a product measure, which gives a weight of $\alpha$ to $A_0$ and of $1-\alpha$ to $A_1$. This comes from the fact that $T$ is affine on each element of the partition, and has full image. I will let you prove that; I think the easiest way is to prove Markov property for the shift and then to prove that the transition matrix is trivial, but I may be missing a simpler argument. Once you have proved that those two systems are conjugated, you have almost won : their entropies are the same, and computing the entropy of a Bernoulli shift is straightforward.

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Ah, I get it now. Thank you for your reply! –  Ivah Jan 18 '12 at 17:50
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