Editing Bernoulli process (section)

=== The 2x mod 1 map===
[[Image:Exampleergodicmap.svg|thumb|The map ''T'' : [0,1) → [0,1), <math>x \mapsto 2x \bmod 1</math> preserves the [[Lebesgue measure]].]]

The above can be made more precise. Given an infinite string of binary digits <math>b_0, b_1, \cdots</math> write
:<math>y=\sum_{n=0}^\infty \frac{b_n}{2^{n+1}}.</math>
The resulting <math>y</math> is a real number in the unit interval <math>0\le y\le 1.</math> The shift <math>T</math> induces a [[homomorphism]], also called <math>T</math>, on the unit interval. Since <math>T(b_0, b_1, b_2, \cdots) = (b_1, b_2, \cdots),</math> one can see that <math>T(y)=2y\bmod 1.</math>  This map is called the [[dyadic transformation]]; for the doubly-infinite sequence of bits <math>\Omega=2^\mathbb{Z},</math> the induced homomorphism is the [[Baker's map]].

Consider now the space of functions in <math>y</math>. Given some <math>f(y)</math> one can find that
:<math>\left[\mathcal{L}_T f\right](y) = \frac{1}{2}f\left(\frac{y}{2}\right)+\frac{1}{2}f\left(\frac{y+1}{2}\right)</math>

Restricting the action of the operator <math>\mathcal{L}_T</math> to functions that are on polynomials, one finds that it has a [[discrete spectrum]] given by 
:<math>\mathcal{L}_T B_n= 2^{-n}B_n</math>
where the <math>B_n</math> are the [[Bernoulli polynomials]]. Indeed, the Bernoulli polynomials obey the identity
:<math>\frac{1}{2}B_n\left(\frac{y}{2}\right)+\frac{1}{2}B_n\left(\frac{y+1}{2}\right) = 2^{-n}B_n(y)</math>