# Correlation between two sequences of irrational numbers

Let us consider the sequence $$x(n+1) = \{b+x(n)\}$$ with $$x(0) = 0$$. Here the brackets represent the fractional part function. Thus $$x(n)= \{nb\}$$ is related to Beatty sequences. If $$b$$ is irrational, it is known that the numbers $$x(n)$$ are uniformly distributed, with a lag-$$k$$ auto-correlation equal to

$$6(b+k)^2 + 6(1-2\lfloor b+k+1\rfloor)(b+k)+6\lfloor b+k+1\rfloor^2 -6\lfloor b+k+1\rfloor+1$$.

This result follows from section 5.4 in this article as well as results published here. Also, if $$b_1$$ and $$b_2$$ are two irrational numbers that are linearly independent over the set of rational numbers, then the correlation between the two sequences (one generated with $$b_1$$ and the other one with $$b_2$$) is zero.

But what if $$b_1 = -1 + \sqrt{5}/2$$ and $$b_2 = 2/\sqrt{5}$$? In that case, I know with absolute certainty (yet with no proof so far) that the correlation between the two sequences is 1/20 = 0.05. How do you prove this result?

Look at the closed subgroup $$G\subset \mathbb T^2$$ (using the notation that $$\mathbb T=\mathbb R / \mathbb Z$$) generated by $$(b_1,b_2)$$; it will be of the form $$G=\{ (ax,bx):x\in\mathbb T\}$$ for certain integers $$a$$ and $$b$$. (In your case, $$(a,b)=(5,4)$$.) Your orbit $$\{(nb_ 1,nb_2):n=1,2,\ldots\}$$ is uniformly distributed in $$G$$. Your correlation is obtained by integrating with respect to the uniform (Haar) measure on $$G$$, namely by $$\int_0^1 (ax\bmod1)(bx\bmod1)dx$$.
It might help in general to develop the Fourier series for the functions $$x\mapsto ax\bmod1$$ and $$bx\mapsto x\bmod1$$ on $$\mathbb T$$.
In your specific case, however, the integral $$I= \int_0^1 (5x\bmod1)(4x\bmod1)dx$$ is given by the following edge-of-tedious expression $$I=\int_0^{1/5}(5x)(4x)dx+ \int_{1/5}^{1/4}(5x-1)(4x)dx+ \int_{1/4}^{2/5}(5x-1)(4x-1)dx+ \int_{2/5}^{1/2}(5x-2)(4x-1)dx+ \int_{1/2}^{3/5}(5x-2)(4x-2)dx+ \int_{3/5}^{3/4}(5x-3)(4x-2)dx+ \int_{3/4}^{4/5}(5x-3)(4x-3)dx+ \int_{4/5}^{1}(5x-4)(4x-3)dx.$$ The correlation coefficient is $$\frac{I-1/4}{1/12}$$.
• I meant $-1+(\sqrt{5}/2)$ but in both cases the correlation will be different from zero and the way to compute it is the same. Commented May 3, 2019 at 15:52
• I think the solution is $-3+ 12\cdot \lim_{T\rightarrow \infty} \frac{1}{T}\int_0^T \{b_1 x\} \{b_2 x\} dx$, but then you need to compute that integral. I'll check if WolframAlpha can do it. Commented May 3, 2019 at 15:55
• Thank you, but it's really just an application of the en.wikipedia.org/wiki/… for the particular Riemann integrable function $x\mapsto(5x\mod1)(4x\mod1)$ applied to the equidistributed sequence $n/\sqrt{20}$. Commented May 4, 2019 at 13:36