2
$\begingroup$

Let $I$ be a (nonempty) compact interval in $\mathbb R$ and $a_1,b_1,\ldots,a_L,b_L \in \mathbb R$. Let $\varphi$ be a piecewise function with $T \ge 2$ pieces(for example $T=2$ for the choice $\varphi(t) = \max(0, t)$ for $t\in \mathbb R$). Consider the piecewise-linear function $f:I \rightarrow \mathbb R$ defined by $f(x)=z_L$, where $$ \begin{split} z_0&=x,\\ z_l &= \varphi(a_l z_{l-1} + b_l),\;\forall l=1,2,\ldots,L. \end{split} $$

Let $a_1,\ldots,a_L$ be drawn i.i.d from some $\mathcal N(0,\sigma_a^2)$ and $b_1,\ldots,b_L$ be drawn i.i.d from some $\mathcal N(0,\sigma_b^2)$.

I'm interested in the distribution of the number of linear pieces of $f$.

Questions

  • What is a good estimate of the average number of linear pieces of $\mathcal f$ as a function of $T$, $L$, $\sigma_a^2$, $\sigma_b^2$, and the length of $I$ ?
  • What kinds of techniques can be used for such problems ?
$\endgroup$
  • $\begingroup$ Doesn't it just depend on the signs of a and b whether the number of linear pieces doubles or not? $\endgroup$ – Dirk Jun 13 at 17:41
  • $\begingroup$ @Dirk Please could be more explicit on your remark ? :) $\endgroup$ – dohmatob Jun 13 at 17:54
  • 1
    $\begingroup$ The next $z_l$ is always zero on some intervall. If the next $b$ is negative while $a$ is positive you'll have one more kink. (However, it occurs to me that the number of kinks may even double every layer, so probably I should stop thinking without a piece of paper.) $\endgroup$ – Dirk Jun 13 at 18:32
  • $\begingroup$ @Dirk Thanks for the input. Such arguments can be used to obtain upper bounds on the max number of pieces, but I don't think they can go beyond this to give insight into means, etc. No ? I have found a result of type Kac-Rice, which from afar at least, should give me something. Indeed, see Theorem 4.1.1 of "Random Fields and their Geometry" pdfs.semanticscholar.org/cb22/…. I will write up what I arrive at, asap. $\endgroup$ – dohmatob Jun 24 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.